Abstract algebra

Section 14.7.

Today we discuss the insolvability of a general quantic in radicals.

Definition 1. An extension K/F is cyclic if it is Galois with cyclic Galois group.

Proposition 2. Let F be a field of characteristic not dividing n, which contains a primitive n-th root of unity. Then for any a ∈ F , the extension F( n

√ a)/F is a cyclic extension of

degree dividing n, here n √ a denotes any root of the polynomial xn −a.

Proof. Since all roots of xn−a can be obtained by multiplying n √ a by n-th roots of unity all

of which lie in F , we see that F( n √ a)/F is the splitting field for xn −a and hence is Galois.

Since the Galois group G = Gal(F( n √ a)/F) permutes the roots of xn −a, for any σ ∈ G we

have σ( n √ a) = ζσ

n √ a for certain n-th root of unity ζσ. Therefore, we get a map

ψ : G −→ Z/nZ, σ 7−→ ζσ, and we leave it as an exercise to show that ψ is a homomorphism. The kernel of ψ consists of those elements which fix n

√ a and is therefore trivial. This shows that ψ is an embedding,

and hence the order of G divides n. �

Definition 3. A character χ of a group G with values in a field L is a homomorphism

χ: G −→ L×.

Definition 4. Character χ1, . . . ,χn of a group G with values in L are linearly dependent if there exist elements a1, . . . ,an ∈ L such that

a1χ1(g) + · · · + anχn(g) = 0 for all g ∈ G, and linearly independent otherwise.

Proposition 5. Distinct characters of a group G with values in a field L are linearly inde- pendent.

Proof. Suppose that characters χ1, . . . ,χm are linearly dependent. Choose the depen- dence relation with the minimal number n of coefficients ai. Let g0 ∈ G be such that χ1(g0) 6= χn(g0), it exists since χ1 6= χn. Then we have

n∑ j=1

ajχj(g0g) = n∑ j=1

ajχj(g0)χj(g) = 0.

Subtracting expression

χn(g0) n∑ j=1

ajχj(g) = n∑ j=1

ajχn(g0)χj(g) = 0

from the left hand side of the previous equality, we get

n−1∑ j=1

aj (χj(g0) −χn(g0)) χj(g) = 0 for all g ∈ G,

which contradicts our assumption that n was the minimal number of non-zero coefficients in a dependence relation for characters χ1, . . . ,χm. �

Proposition 6. Let K/F be a cyclic extension of degree n, where the field F contains a primitive n-th root of unity and char(F) does not divide n. Then K = F( n

√ a) for some

a ∈ F . 1

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Proof. Let σ ∈ G = Gal(K/F) be the generator of the cyclic Galois group. For any element α ∈ K and any n-th root of unity ζ, consider the Lagrange resolvent

(α,ζ) =

n−1∑ k=0

ζkσk(α) = α + ζσ(α) + · · · + ζn−1σn−1(α).

Since ζ ∈ F and therefore σ(ζ) = ζ, we have

σ(α,ζ) = n−1∑ k=0

ζkσk+1(α) = ζ−1 n−1∑ k=0

ζk+1σk+1(α) = ζ−1 n∑ k=1

ζkσk(α) = ζ−1(α,ζ),

where the last equality follows from the fact that ζn = 1 ∈ F and σn = 1 ∈ G. Now, we see that

σ ((α,ζ)n) = ( ζ−1 )n

(α,ζ)n = (α,ζ)n,

which implies (α,ζ)n ∈ F for any α ∈ K. Let us now choose such an α ∈ K that (α,ζ) 6= 0, which exists because the characters

1,σ, . . . ,σn−1 are independent. Let us also pick ζ to be a primitive root of unity. Then

σk(α,ζ) = ζ−k(α,ζ),

which means that (α,ζ) ∈ K is not fixed by any nontrivial element of G, and therefore does not lie in any proper subfield of K. We now conclude that K = F ((α,ζ)), and since (α,ζ)n ∈ F , we can write K = F( n

√ a) for a = (α,ζ)n. �

In what follows, for simplicity we will assume char(F) = 0. Nonetheless, all the results will be valid for the fields whose characteristic does not divide orders of the roots taken.

Definition 7.

(1) Let K = F( n √ a) for some element a ∈ F. Then K/F is a simple radical extension.

(2) A field K is a root extension of a field F if there exists a chain of subfields

F = K0 ⊂ K1 ⊂ ···⊂ Ks = K,

such that Ki/Ki−1 is s simple radical extension for all 1 = 1, . . . ,s. (3) An element α, which is algebraic over F , can be expressed by radicals if α ∈ K for

some root extension K/F. (4) A polynomial f(x) ∈ F[x] can be solved by radicals if all of its roots can be expressed

by radicals.

Remark 8. Simply speaking, a polynomial can be solved by radicals over a field F, if all of its roots can be obtained by applying successive operations of addition, subtraction, multiplication, division, and extraction of n-th roots to elements of F .

Lemma 9. For every root extension K/F , there exists a field E ⊃ K, such that E/F is a Galois root extension, and each simple radical extension Ei/Ei−1 is cyclic.

Proof. Let L be the Galois closure of K. Then for any σ ∈ Gal(L/F), there exists a chain of subfields

F = σ(K0) ⊂ σ(K1) ⊂ ···⊂ σ(Ks) = σ(K), where σ(Ki/Ki−1) is again a simple radical extension:

Ki = Ki−1 ( ni √ ai) implies σ(Ki) = σ(Ki−1)

( ni √ σ(ai)

) .

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Note also, that a composite of two root extensions is also a root extension. Indeed, if K/F and K′/F are root extensions with intermediate extensions Ki and K

′ j respectively, one can

consider a chain

F = K0 ⊂ ···⊂ Ks = K = KK′0 ⊂ KK ′ 1 ⊂ KK

′ r = KK

′.

It follows that the composite of all fields of the form σ(K), σ ∈ Gal(L/F), is again a root extension, but the latter composite is precisely L.

Now we have a Galois root extension L/F with a chain of subfields

F = L0 ⊂ L1 ⊂ ···⊂ Lk = L, where Li = Li−1( ni

√ ai) for all i = 1, . . . ,k. Choose a number n such that ni |n for all

1 6 i 6 k, and set F ′ = F(ζn), where ζn is a primitive n-th root of unity. Then the extension F ′/F is a simple radical extension. Finally, consider the composition E = LF ′, which is Galois over F since both L and F ′ are. Moreover, we have the chain of subfields

F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = E, and each extension F ′Ki/F

′Ki−1 is cyclic by the first proposition in this lecture. �

Recall that a group G is solvable if there exists a chain of normal subgroups

1 = Gs CGs−1 C · · ·CG1 CG0 = G, with cyclic quotient groups Gi−1/Gi for all i = 1, . . . ,s. Recall also that subgroups and quotient groups of a solvable group are themselves solvable, and that if both H ⊂ G and G/H are solvable, then so is G.

Theorem 10. A polynomial can be solved in radicals if and only if its Galois group is solvable.

Proof. Suppose that f(x) ∈ F [x] can be solved in radicals. Then each root of f(x) is con- tained in an extension as in the lemma above. Consider the composite L of these extensions for all roots of f(x). Then L/F is again a Galois root extension with intermediate cyclic extensions Li/Li−1. Now, let G = Gal(L/F) and Gi be the subgroups of G corresponding to the subfields Li. Since

Gal(Li/Li−1) = Gi−1/Gi,

we see that G is solvable. Moreover, since L contains the splitting field of the polynomial f(x), the Galois group H of f(x) is a subgroup of G and therefore solvable.

Suppose now that the Galois group H of f(x) is solvable, and let K be the splitting field of f(x). Consider the chain of subgroups

1 = Hs CHs−1 C · · ·CH1 CH0 = H, with cyclic quotient groups Hi−1/Hi, and let Ki = K

Hi be the fixed fields of these subgroups. Then we get a chain of cyclic extensions

F = K0 ⊂ K1 ⊂ ···⊂ Ks = K. Considering an extension F ′/F as in the preceding lemma, we have a chain of cyclic extensions

F ⊂ F ′ = F ′K0 ⊂ F ′K1 ⊂ ···⊂ F ′Ks = F ′K. Now however, each extension F ′Ki/F

′Ki−1 is a simple radical extension by one of the Propo- sitions we proved today, since F ′Ki−1 contains the appropriate roots of unity. Finally, since all roots of f(x) are contained in F ′K, we see that f(x) can be solved in radicals. �

Corollary 11. The general equation of degree > 5 is not solvable in radicals.

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Proof. The Galois group of a general polynomial of degree n is isomorphic to Sn, and Sn is not solvable for n > 5. To show that Sn is not solvable, it is enough to notice that the subgroup An ⊂ Sn is simple for n > 5. The latter follows from the following easy to shoe facts: An is generated by 3-cycles for n > 5, any normal subgroup of An contains a 3-cycle, hence contains all 3-cycles. �