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STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER
FIELDS
1 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1) over C.
a) Find the dimension of Aas a C-vector space.
b) Determine the irreducible representations of Aover C.
Solution 1.
a) The dimension of Aas a C-vector space is equal to the dimension of the algebra modulo
the ideal (x2−1). Note that Ais isomorphic to the ring of polynomials of degree at most 1 over
C, denoted C[x]/(x2−1) ∼
=C[x]/(x−1)(x+ 1). This is isomorphic to C⊕Cas a vector space.
Therefore, the dimension of Ais 2.
b) To determine the irreducible representations of Aover C, we need to find the distinct rep-
resentations of Aas a direct sum of simple submodules. Since Ahas dimension 2, all irreducible
representations of Aare 1-dimensional. We consider two possible irreducible representations:
- The trivial representation, where the action of Aon any vector in the basis of C⊕Cis the
identity map.
- The non-trivial representation, where the action of Ais given by sending xto −1(or equiva-
lently, sending xto 1).
These are the two distinct irreducible representations of Aover C.
2 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra Aover the field Rdefined by the multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Show that Ais a non-commutative algebra.
b) Determine if Ais a simple algebra over R.
Solution 1. a) To show that Ais non-commutative, we need to find elements x, y ∈Asuch
that x·y=y·x. Let’s take x=aand y=b:
a·b=ebut b·a=b, so Ais indeed non-commutative.
b) To determine if Ais simple, we need to check if it has any non-trivial two-sided ideals. We
will show that the only two-sided ideals of Aare {e}and Aitself, which implies that Ais simple.
Let Ibe a nonzero two-sided ideal of A. Since Ais finite and non-commutative, it follows that
Ais a non-commutative division algebra. Thus, Imust be a non-trivial subalgebra of A.
Consider the non-zero element ain I. Since Ais a division algebra, it must have an inverse
a−1. Since Iis a two-sided ideal, we have a·(a−1·a)=(a·a−1)·a∈I. But a·a−1=e, so
a·e=a∈I.
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
Similarly, b∈Ias well. By the multiplication table, a·b=eand b·a=b, so e∈Ias well.
Therefore, Icontains all elements of A, meaning I=A. This shows that Ahas no non-trivial
two-sided ideals, and therefore is a simple algebra over R.
3 PRIME AND MAXIMAL IDEALS IN ASSOCIATIVE ALGEBRAS: CLASSIFICATION AND
PROPERTIES
Problem 4. Let kbe a field and consider the associative algebra A=k[x]/(x2), where k[x]is
the polynomial ring in one variable over k. Determine the prime and maximal ideals in A.
Solution 4.
Given the associative algebra A=k[x]/(x2), we first note that Ais isomorphic to the ring of
dual numbers over k. The elements of Acan be written as a+bx, where a, b ∈k.
a) To determine the prime ideals in A, we observe that Ais a local ring with maximal ideal (x),
since every non-unit in Ais divisible by x. Thus, (x)is the only maximal ideal in A, and it is also a
prime ideal.
b) Now, let I⊂Abe an ideal. We want to find the prime ideals that contain I. Since Ais a
principal ideal domain, any ideal in Ais of the form (a+bx)for some a, b ∈k.
If I= 0, then (0) ⊂Ais a prime ideal as it is included in the maximal ideal (x).
If I= (a+bx)where a= 0, then Iis a prime ideal if and only if ais prime in k. This is because
any zero divisors in klift to zero divisors in A, which would prevent the ideal from being prime.
c) The maximal ideals in Aare precisely those of the form (x−α)where α∈k. This follows
from the fact that (x)is the only maximal ideal in A, and each maximal ideal must be of the form
(x−α)for some α∈k.
4 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 4. Consider an associative algebra Aover a field Fwith the following multiplication
table:
·e a b
e e a b
a a e b
b b b e
a) Verify whether Ais a division algebra.
b) Find the center of A.
c) Determine whether Ais simple.
Solution 4.
a) To check if Ais a division algebra, we need to see if every non-zero element in Ahas a
multiplicative inverse. In this case, let’s consider the element a. The product aa =ewhich is the
identity element. So, adoes have a multiplicative inverse. Similarly, we can see that band ealso
have multiplicative inverses. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A. Let
cbe an element in the center of A. Then we have the conditions ce =ec,ca =ac, and cb =bc.
From the multiplication table, we can see that cmust be equal to e. Therefore, the center of Ais
{e}.
c) An algebra Ais simple if its only two-sided ideals are {0}and Aitself. Let Ibe a two-sided
ideal of A. From the multiplication table, we can see that the ideal Imust contain aand b, but then
it would also have to contain eby closure under multiplication. Therefore, the only two-sided ideals
of Aare {0}and A. Hence, Ais simple.
5 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=R[x]/(x2−1) over the field of real numbers.
a) Determine the dimension of Aas a vector space over R.
b) Find the irreducible representations of A.
Solution 1. a) Since A=R[x]/(x2−1), we can write A={a+bx |a, b ∈R}. Thus, Ais a
2-dimensional vector space over R.
b) To find the irreducible representations of A, we first note that R[x]/(x2−1) splits into two
components: R[√2] and R[−√2].
The irreducible representations of Aare given by the distinct field automorphisms of the field
extension R[√2]/R. Since √2is a square root of 2, the distinct field automorphisms are the identity
map and the map that sends √2to −√2.
Thus, the irreducible representations of Aare the trivial representation and the sign represen-
tation (which changes the sign of the basis vector corresponding to √2).
Problem 2. Let B=C[x]/(x3−1) be an associative algebra over the field of complex numbers
C.
a) Determine the dimension of Bas a vector space over C.
b) Find the irreducible representations of B.
Solution 2. a) Since B=C[x]/(x3−1), we can write B={a+bx +cx2|a, b, c ∈C}. Thus,
Bis a 3-dimensional vector space over C.
b) The ring C[x]/(x3−1) can be decomposed into three components: C,C[ω], and C[ω2],
where ωis a primitive cube root of unity.
The irreducible representations of Bcorrespond to the distinct field automorphisms of the field
extension C[ω]/C. These automorphisms are the identity map, the map that sends ωto ω2, and
the map that sends ωto ω4=ω2.
Thus, the irreducible representations of Bare the trivial representation, the standard 3-dimensional
representation of C, and the conjugate representation of the standard representation.
6 "CLASSIFICATION AND PROPERTIES OF SIMPLE ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 6. Let Abe a simple associative algebra over a field Fwith dimension 4. Determine
the possible isomorphism classes of A.
Solution 6.
a) Since Ais a simple algebra, it has no nontrivial proper ideals. By the Artin-Wedderburn
theorem, Ais isomorphic to a matrix algebra over a division ring. Since Ais of dimension 4, the
only possible division ring is the field Fitself. Thus, Ais isomorphic to M2(F), the algebra of 2×2
matrices over F.
b) Now we need to determine the non-isomorphic simple algebras of the form M2(F). The
isomorphism classes of M2(F)are determined by the field F.
c) Therefore, the possible isomorphism classes of Aare given by isomorphisms to M2(F)where
Franges over all possible fields. In particular, when Fis R,C, or a finite field Fp, the corresponding
isomorphism classes of Aare Mat2(R), Mat2(C), and Mat2(Fp), respectively.
7 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 1. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Awith dimension 4over K. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, it has no nontrivial two-sided ideals. By the Artin-
Wedderburn Theorem, any simple algebra over an algebraically closed field Kis isomorphic to a
matrix algebra over K.
Since Ais of dimension 4, it is isomorphic to M2(K), the 2×2matrix algebra over K. So, the
possible isomorphism classes of Aconsist of only one element, which is M2(K).
Problem 2. Let Kbe an algebraically closed field and consider the simple associative K-
algebra Bwith dimension 9over K. Determine the possible isomorphism classes of B.
Solution 2. Similar to the previous problem, since Bis a simple algebra, it is isomorphic to a
matrix algebra over Kaccording to the Artin-Wedderburn Theorem.
For a 9-dimensional algebra, we consider the possibilities. The options for the dimension of a
matrix algebra over Kcan be 1×1,2×2,3×3, and 3×3which gives dimension 9.
Therefore, the possible isomorphism classes of Bare M3(K), the 3×3matrix algebra over K.
8 MINIMAL GENERATING SETS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field Fwith basis {1, x, x2}and multiplication
given by x3=x2+ 2x. Determine if the set {1 + x, x +x2}is a minimal generating set for A.
Solution 1.
To determine if the set {1 + x, x +x2}is a minimal generating set for A, we need to check
whether each element of the basis can be obtained as a linear combination of elements in this set.
a) 1 = (1 + x)−x, so 1can be obtained using the elements of the set.
b) x=x+x2−x2=x+x2, so xcan be obtained using the elements of the set.
c) x2= (x+x2)−x=x2, so x2can be obtained using the elements of the set.
Since all basis elements can be generated using the set {1+x, x+x2}, it is indeed a generating
set for A. To check for minimality, we need to verify if any proper subset generates A. However, as
the set contains only 2 elements and removing any one of them would result in a set that cannot
generate x, the set {1 + x, x +x2}is a minimal generating set for A.
9 IRREDUCIBLE REPRESENTATIONS OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 9. Consider the associative algebra A=R[x]/(x2−1) over the field R. Find the
irreducible representations of A.
Solution 9. a) To find the irreducible representations of A=R[x]/(x2−1), we first determine
the possible values for xin A. Since x2−1=0, we have that x2= 1, implying that x=±1. Thus,
the only possible eigenvalues for xare λ= 1 and λ=−1.
b) Next, we find the corresponding eigenvectors for x= 1 and x=−1.
For x= 1, we solve for the eigenvectors of xwith eigenvalue λ= 1:
(x−1)v= 0 =⇒(1 −1)v= 0 =⇒0v= 0
Thus, any non-zero vector can be an eigenvector for x= 1.
For x=−1, we solve for the eigenvectors of xwith eigenvalue λ=−1:
(x+ 1)v= 0 =⇒(−1 + 1)v= 0 =⇒0v= 0
Again, any non-zero vector can be an eigenvector for x=−1.
c) Since we have found that any non-zero vector could be an eigenvector for x= 1 and x=−1,
we have infinite one-dimensional representations of A. Thus, the irreducible representations of A
are all one-dimensional representations with eigenvalues λ= 1 or λ=−1.
10 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Consider the associative algebra A=C[x]/(x2−1), where Cis the field of complex
numbers. Determine whether Ais simple, and if not, find a nontrivial two-sided ideal of A.
Solution 1. To determine if Ais simple, we will first find all the two-sided ideals in A.
Since A=C[x]/(x2−1), any element in Acan be written as a+bx + (x2−1) for a, b ∈
C. Let’s consider a general form of an ideal Iin Agenerated by p(x) + (x2−1), where p(x)
is an arbitrary polynomial in C[x]. We want to find all elements q(x)+(x2−1) ∈Asuch that
(q(x)+(x2−1))(p(x)+(x2−1)) ∈I. Expanding this, we get:
(q(x)+(x2−1))(p(x)+(x2−1)) = q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2.
For the result to be in I, we must have q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) + (x2−1)2= 0,
which implies that q(x)p(x) + q(x)(x2−1) + p(x)(x2−1) = 0. From this, we see that the ideal Iis
generated by all multiples of x2−1.
Since Ahas a nontrivial two-sided ideal, it is not a simple algebra. The ideal generated by x2−1
is a nontrivial two-sided ideal of A.
Therefore, the algebra A=C[x]/(x2−1) is not simple and has a nontrivial two-sided ideal
generated by x2−1.
11 "STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 12. Let Abe a simple associative algebra over an algebraically closed field with
dimension 4as a vector space. If Ais not isomorphic to the algebra of 2×2matrices over the field,
determine the possible isomorphism class(es) of A.
Solution 12. Given that Ais a simple associative algebra of dimension 4over an algebraically
closed field, and it is not isomorphic to the algebra of 2×2matrices, we need to find the possible
isomorphism class(es) of A.
Since Ais simple, its only simple ideals are {0}and Aitself. The Schur’s lemma states that if
an algebra Aover an algebraically closed field does not have any nontrivial two-sided ideals, then
any two simple modules over Aare isomorphic.
Now, if Ais not isomorphic to the algebra of 2×2matrices, the dimension of the matrix algebra
over the field would be 4. But the dimension of the matrices of size 2×2is 4, so Amust be one of
the other simple algebras of dimension 4.
The possible isomorphism class(es) of Aare: 1. M2(F)(the algebra of 2×2matrices over F),
2. The quaternion algebra H(F), 3. The octonion algebra O(F).
Therefore, Acould be isomorphic to the quaternion algebra or the octonion algebra, apart from
being isomorphic to the 2×2matrices.
12 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let F=Z2be the field with two elements. Consider the associative algebra A
over Fdefined by the multiplication table:
·0 1
0 0 0
1 0 1
a) Show that Ais a division algebra.
b) Determine the center of A.
Solution 1.
a) To show that Ais a division algebra, we need to verify that every non-zero element has a
multiplicative inverse. In this algebra, both elements 0and 1have inverses: 0is its own inverse,
and 1is also its own inverse. Therefore, Ais a division algebra.
b) The center of an algebra Ais the set of elements that commute with every element in A.
For Awith the given multiplication table, we see that 0commutes with all elements and 1only
commutes with itself. Therefore, the center of Ais {0}.
13 PRIMITIVE IDEMPOTENT ELEMENTS IN ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 13. Consider the associative algebra Aover the field Rwith basis {1, x, x2, x3}and
multiplication table given by:
1x x2x3
1 1 x x2x3
x x x2x31
x2x2x31x
x3x31x x2
a) Find all primitive idempotent elements of A.
b) Determine the dimension of the centralizer of each primitive idempotent.
c) Classify the simple modules of Aup to isomorphism.
Solution 13.
a) To find the primitive idempotent elements of A, we need to look for idempotents e∈Asuch
that e2=eand they cannot be written as a sum of two orthogonal idempotents.
From the multiplication table, we observe that the idempotent elements of Aare 1and x2, as
(x2)2=x3=x2and (1)2= 1. Since x2cannot be written as a sum of orthogonal idempotents, it is
the only primitive idempotent of A.
b) To determine the dimension of the centralizer of x2, we look for elements c∈Asuch that
cx2=x2c. The centralizer of x2can be represented as a linear combination of 1and x2, as the
other elements do not commute with x2. Thus, the dimension of the centralizer of x2is 2.
c) Since x2is the only primitive idempotent element of A, the simple modules of Aup to iso-
morphism are all the one-dimensional vector spaces spanned by x2, given by {a(x2)|a∈R}.
Therefore, the simple modules of Aare the one-dimensional vector spaces spanned by x2, up
to isomorphism.
14 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe a simple Artinian algebra over an algebraically closed field kof dimension
3. Determine the possible isomorphism classes of A.
Solution 1. Given that Ais a simple Artinian algebra of dimension 3over an algebraically
closed field k, we know that there are three possibilities for the isomorphism class of A:
a) If Ais isomorphic to M3(k), the 3×3matrix algebra over k.
b) If Ais isomorphic to M2(k)⊕k, the direct sum of 2×2matrices over kwith a 1-dimensional
simple k-module.
c) If Ais isomorphic to k[x]/(f(x)), where f(x)is an irreducible polynomial of degree 3.
We can rule out option c) since the dimension of k[x]/(f(x)) is 3but it is not simple.
Therefore, the possible isomorphism classes of Aare M3(k)and M2(k)⊕k.
15 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 16. Let Abe a simple Artinian algebra over a field Fwith dimFA= 4. Consider the
following elements in A:
v1= (2,3,1,4), v2= (1,2,3,4), v3= (3,2,1,4), v4= (1,3,2,4).
a) Show that the set {v1, v2, v3, v4}is linearly independent.
b) Determine a basis for Aover F.
c) Find the dimension of the center of A.
Solution 16.
a) To show that the set {v1, v2, v3, v4}is linearly independent, we need to show that the only
solution to the equation c1v1+c2v2+c3v3+c4v4= 0 with ci∈Ffor all iis c1=c2=c3=c4= 0.
Setting c1v1+c2v2+c3v3+c4v4= 0 gives:
c1(2,3,1,4) + c2(1,2,3,4) + c3(3,2,1,4) + c4(1,3,2,4) = (0,0,0,0).
This leads to the following system of equations:
2c1+c2+ 3c3+c4= 0
3c1+ 2c2+ 2c3+ 3c4= 0
c1+ 3c2+c3+ 2c4= 0
4c1+ 4c2+ 4c3+ 4c4= 0
.
Solving this system, we find that c1=c2=c3=c4= 0, which proves linear independence.
b) Since dimFA= 4 and {v1, v2, v3, v4}is a linearly independent set, it forms a basis for Aover
F.
c) The dimension of the center of A, denoted as Z(A), satisfies dimFZ(A) = dimFA−
dimF[A, A], where [A, A]is the commutator ideal of A.
The commutator ideal [A, A]is the smallest two-sided ideal in Asuch that A/[A, A]is a simple
algebra over F. Since Ais simple, [A, A]must be the zero ideal. Therefore, dimFZ(A) = dimFA−
dimF[A, A] = 4 −0=4. Hence, the dimension of the center of Ais 4.
Therefore, the basis for Aover Fis {v1, v2, v3, v4}, and the dimension of the center of Ais 4.
16 DECOMPOSITION AND CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS
Problem 16. Let Fbe a field and consider the associative algebra A=F[x]/(x2−2).
a) Show that Ais a simple associative algebra.
b) Determine the center of A.
c) Find the dimension of Aas a vector space over F.
Solution 16.
a) To show that Ais simple, we need to show that Ahas no non-trivial two-sided ideals. By the
First Isomorphism Theorem, Ais isomorphic to F(√2), which is clearly a field. Since fields have
no non-trivial ideals, Ais a simple associative algebra.
b) The center of A, denoted by Z(A), consists of all elements in Athat commute with every
element in A. Since Ais isomorphic to F(√2),Z(A) = F.
c) The dimension of Aas a vector space over Fis equal to the degree of the minimal polynomial
of the algebraic element in A. Here, the algebraic element is √2, and the minimal polynomial is
x2−2. Therefore, dimFA= deg(x2−2) = 2.
17 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Fbe a field and consider the associative algebra A=M2(F)of 2×2matrices
with entries from F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided
ideals.
Problem 2. Let Fbe a field and consider the associative algebra A=F[x]of polynomials over
F. Determine if Ais simple or not, and if not simple, find its nontrivial two-sided ideals.
Problem 3. Let Fbe a field and consider the associative algebra A=F×Fwith component-
wise addition and multiplication. Determine if Ais simple or not, and if not simple, find its nontrivial
two-sided ideals.
18 SOLUTION
Solution 1. The algebra A=M2(F)is not simple. To find its nontrivial two-sided ideals, we note
that the set of all scalar matrices is an ideal of A. Furthermore, the set of all matrices of the form
0
a
0
where a∈Fis also an ideal of A. Therefore, Ahas at least two nontrivial two-sided
ideals, namely the set of scalar matrices and the set of diagonal matrices as described.
Solution 2. The algebra A=F[x]is not simple. Any polynomial p(x)∈F[x]generates an
ideal of Aconsisting of all multiples of p(x). Therefore, Ahas infinitely many nontrivial two-sided
ideals generated by nonconstant polynomials.
Solution 3. The algebra A=F×Fis not simple. The zero ideal {(0,0)}and the full algebra
Aare both nontrivial two-sided ideals of A. Thus, Ahas at least two nontrivial two-sided ideals, as
shown.
19 “STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS"
Problem 19. Let Abe a simple associative algebra over a field F, with dimF(A)=4. Suppose
that Ais not isomorphic to the matrix algebra M2(F). Determine the possible structure of A.
Solution 19. Since Ais not isomorphic to M2(F), we know that Acannot be isomorphic to a
direct sum of two copies of F(i.e., AF ⊕F⊕F⊕F).
Therefore, the only possibility is that Ais a division algebra over F(i.e., Ais a division ring). This
implies that Amust be a field, as any nonzero element in a division ring must have a multiplicative
inverse.
So, the possible structure of Ain this case is that Ais a field extension of F, where dimF(A)=4.
20 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
Problem 1. Let Abe an associative algebra over a field F, and let a, b ∈Abe nilpotent
elements. Suppose aand bcommute, i.e., ab =ba. Show that a+bis also a nilpotent element in
A.
Solution 1. Since aand bare nilpotent elements, there exist positive integers mand nsuch
that am= 0 and bn= 0.
Now consider the element c=a+b. We aim to show that cis nilpotent by finding a positive
integer ksuch that ck= 0.
Using the binomial theorem, we have
c2= (a+b)2=a2+ab +ba +b2=a2+ 2ab +b2
=a2+ 2ba +b2[Since ab =ba]
=a2+ 2ab +b2=a2+b2[Since a2= 0, b2= 0].
Similarly, we can show:
c3=a3+b3= 0 [Since a3= 0, b3= 0].
Thus, we have found a positive integer k= 3 such that ck= 0, which means c=a+bis a
nilpotent element in A.
Therefore, if aand bare nilpotent elements that commute, then their sum a+bis also a nilpotent
element in A.
21 STRUCTURE THEORY OF ASSOCIATIVE ALGEBRAS OVER FIELDS
21.1 THE CLASSIFICATION OF SIMPLE ASSOCIATIVE ALGEBRAS OVER ALGE-
BRAICALLY CLOSED FIELDS
Problem 1. Let Kbe an algebraically closed field and Abe a simple associative algebra over K
with dimension 12. Determine the possible isomorphism classes of A.
Solution 1. Since Ais a simple algebra, its only two-sided ideals are {0}and Aitself. By the
Artin-Wedderburn theorem, Ais isomorphic to a matrix algebra over a certain division ring.
Since Ahas dimension 12 over K, it can only be isomorphic to M3(K)(the 3×3matrices over
K) or M2(D)(the 2×2matrices over a division ring Dwith dimension 6over K).
Therefore, the possible isomorphism classes of Aare M3(K)and M2(D).
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