Abstract algebra

Section 14.3.

In this lecture we will focus on finite fields. Let us collect everything we learnt about them so far.

• Every finite field is isomorphic to Fpn for some prime p and some integer n > 1. • The field Fpn is the splitting field for the polynomial xp

n−x ∈ Fp[x]. Its Galois group is isomorphic to Z/nZ and is generated by the Frobenius automorphism σp : α 7→ αp. • The subfields of Fpn are in bijection with divisors d of n, divisor d corresponds to the

subfield fixed by σdp , this subfield is isomorphic to Fpd . Every subfield of Fpn is Galois over Fp.

The latter point follows from the Fundamental Theorem of Galois Theory together with an observation that Z/nZ is commutative, and hence every subgroup thereof is normal. To demonstrate some applications of the above, let us prove the following easy corollary.

Corollary 1. The irreducible polynomial x4 + 1 ∈ Z[x] is reducible modulo every prime. Proof. If p = 2, we simply have x4 + 1 = (x+ 1)4 in F2[x]. If p is odd, then p is congruent to ±1 or ±3 modulo 8, and hence p2 ≡ 1 mod 8. Then we have (x8 − 1) | (xp2−1 − 1), which implies that (x4+1) |xp2−x. Therefore, every root of x4+1 is also a root of xp2−x. Roots of the latter polynomial form the field Fp2 , and hence any root of x4 + 1 generates an extension of degree at most 2 over F2. Now, we see that x4 + 1 is reducible over Fp, because otherwise it would generate an extension of degree 4. �

Corollary 2. Field Fpn is a simple extension of Fp. In particular, for every n > 1, there exists an irreducible polynomial of degree n over Fp. Proof. First, note that the statement follows from the fact that the multiplicative group F×pn is cyclic. Indeed, if θ is the generator of F×pn , then Fpn ' Fp(θ). Now, since F×pn is an abelian group, we can write it as

F×pn ' Z/a1Z× · · · × Z/akZ, where a1 | a2 | . . . | ak. If k > 1, F×pn would contain more than a1 elements whose order divides a1, and hence the polynomial x

a1 − 1 would have more than a1 roots in Fpn , which leads to a contradiction. �

Corollary 3. The polynomial xp n −x is the product of all distinct irreducible polynomials in

Fp[x] of degree d, where d runs over all divisors of n. Proof. We need to show the following two statements: first, every irreducible polynomial p(x) ∈ Fp[x] of degree d |n divides xp

n − x, and second, if d is the degree of an irreducible polynomial p(x) ∈ Fp[x] which divides xp

n − x, then d |n. Let p(x) ∈ Fp[x] be an irreducible polynomial of degree d |n. Then |Fp[x]/(p(x)) : Fp| = d,

and therefore Fp[x]/(p(x)) ' Fpd . Recall that Fpd/Fp is Galois, hence every root of p(x) is an element of Fpd . At the same time, every element of Fpd is a root of xp

d−x which implies that p(x) | (xpd − x). Now, since d |n we have (xpd − x) | (xpn − x), and therefore p(x) | (xpn − x).

Conversely, let p(x) ∈ Fp[x] be an irreducible polynomial which divides xp n − x, and let

d = deg(p). If p(α) = 0, then Fp ⊂ Fp(α) ⊂ Fpn and |Fp(α) : Fp| = d. The latter implies d |n. �

Note that we can use the above corollary to find products of all irreducible polynomials of a given degree in Fp[x]. For example, product of all linear polynomials over Fp is equal to

x(x− 1) . . . (x− p+ 1) = xp − x. 1

2

Next, product of all quadratics over F2 is equal to x4 − x x(x− 1)

= x2 + x+ 1,

and we conclude that x2 + x + 1 is the unique irreducible polynomial of degree 2 over F2. Similarly, product of all cubics over F2 is equal to

x8 − x x(x− 1)

= x6 + x5 + x4 + x3 + x2 + x+ 1 = (x3 + x+ 1)(x3 + x2 + 1),

and product of all quartics over F2 is equal to x16 − x

x(x− 1)(x2 + x+ 1) = (x4 + x3 + x2 + x+ 1)(x4 + x3 + 1)(x4 + x+ 1).

Let us remark that if p1(x) and p2(x) are irreducible polynomials over Fp of the same degree, say d, then

Fp[x]/(p1(x)) ' Fpd ' Fp[x]/(p2(x)). The polynomial p2(x) splits into linear factors in Fp[x]/(p1(x)), and let α(x) be a root of p2(x) in Fp[x]/(p1(x)). Then the isomorphism Fp[x]/(p1(x)) ' Fp[x]/(p2(x)) can be written explicitly by x 7→ α(x). For example, one can check that x3 + x2 is a root of x4 + x3 + 1 in F2[x]/(x4 + x+ 1), that is (x4 + x+ 1) divides (x4 + x3 + 1)|x 7→x3+x2 . Then

F2[x]/(x4 + x+ 1) ' Fp[x]/(x4 + x3 + 1), x 7→ x3 + x2. Now, we shall prove the Möbius inversion formula and use it to derive the number of

irreducible polynomials of degree n in Fp[x].

Definition 4. The Möbius function µ : Z>0 → Z is defined by

µ(n) =

 1 for n = 1,

0 if n has a square factor,

(−1)r if n has r distinct prime factors.

Consider the set of arithmetic functions A = {f : Z>0 → C} and define a Dirichlet product on A by

(f ◦ g)(n) = ∑

d1d2=n

f(d1)g(d2).

Note that the Dirichlet product is commutative and associative:

((f ◦ g) ◦ h)(n) = (f ◦ (g ◦ h))(n) = ∑

d1d2d3=n

f(d1)g(d2)h(d3).

Then functions

e(n) =

{ 1 n = 1,

0 n > 1, and I(n) ≡ 1

are units for the Dirichlet and the usual multiplication on A respectively. Now, for any f ∈ A define Sf ∈ A by

Sf = I ◦ f = ∑ d |n

f(d).

Let us show that µ is the inverse of I with respect to the Dirichlet product:

I ◦ µ = e.

3

Indeed, let n = pn11 . . . p nr r > 1 be the decomposition of n into prime factors, then

(I ◦ µ)(n) = ∑ d |n

µ(d) = µ(1) +

r∑ k=1

∑ 16i1<···<ik6r

µ(pi1 . . . pik) =

r∑ k=0

( r

k

) (−1)k = 0.

Now, the Möbius inversion formula claims

f(n) = ∑ d |n

µ(d)Sf

(n d

) .

The proof is almost immediate:

µ ◦ Sf = µ ◦ (I ◦ f) = (µ ◦ I) ◦ f = e ◦ f = f. Now let us get back to finite fields. Let ψ(n) be the number of irreducible polynomials of

degree n over Fp. By the last Corollary we have

pn = ∑ d |n

dψ(d).

By the Möbius inversion formula

nψ(n) = ∑ d |n

µ(d)pn/d,

or equivalently,

ψ(n) = 1

n

∑ d |n

µ(d)pn/d.

Finally, let us recall that there is a partial order on fields Fpn , namely Fpm ⊂ Fpn if and only if m |n. Since any finite extension of Fp is isomorphic to Fpn for some n, we see that the algebraic closure of Fp can be written as

Fp = ⋃ n>1

Fpn .