Recall that F∗ is a multiplicative group of order pn− 1

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Dec. 8, 2006

3.8. finite fields. The Galois theory on finite fields is comparatively easy and basically governed by Frobenius map.

Recall that given a finite field F of q elements, it’s prime field must be of the form F_p for some prime p. Let n = [F : F_p], then |F | = pⁿ. Theorem 3.8.1. F is a finite field with pⁿ elements if and only if F is a splitting field of x^pn − x over F_p.

Sketch. Recall that F^∗ is a multiplicative group of order pⁿ− 1. Hence it’s easy to see that every element u ∈ F satisfying x^pn − x. Thus element of F are exactly roots of x^pn − x, therefore, F is a splitting field of x^pn− x over F_p.

Conversely, if F is a splitting field of x^pn− x over F_p. Let E ⊂ F be the subset of all roots of x^pn − x. One can check that E is a subfield (containing F_p and all roots). By definition of splitting field, E is a splitting field, and E = F . So |F | = |E| ≤ pⁿ. However, notice that

x^pn− x is separable. So |F | = pⁿ. ¤

Proposition 3.8.2. Let F be a finite field and F/K is an extension.

Then F/K is Galois. The Galois group is cyclic, generated by Frobenius map.

Proof. We shall prove the case that K = F_p. For general K, F_p ⊂ K ⊂ F . Since F/F_p is Galois, then F/K is also Galois with Galois group K⁰ < Gal_FpF also a cyclic group.

Now we consider F/F_p, and |F | = pⁿ. Since F is a splitting field of a separable polynomial x^pn − x over F_p, F is Galois over F_p.

The Galois group Gal_FpF has order [F : F_p] = n. Consider the Frobenius map ϕ : a → a^p, which is clearly a Fp-automorphism. So ϕ ∈ GalFpF . Note that order of ϕ is n. So GalFpF can only be the

cyclic group generated by ϕ. ¤

3.9. cyclotomic extension. We now start the study of cyclotomic extension.

Definition 3.9.1. A cyclotomic extension of order n over K is a split- ting field of xⁿ− 1.

Remark 3.9.2. If char(K) = p and n = p^rm, then xⁿ−1 = (x^m−1)^pr. Hence we may assume that either char(K) = 0 or char(K) = p - n in the study of cyclotomic extension.

The main theorem is the following:

Theorem 3.9.3. Keep the notation as above. Then we have (1) F = K(ζ), where ζ is a primitive n-th root of unity.

(2) F/K is Galois whose Galois group Gal_F/K can be identified as a subgroup of Z^∗_n.

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(3) If n is prime, then GalF/K is cyclic. More general, is n = p^k with p 6= 2, then then GalF/K is cyclic.

Proof. Let S := {u ∈ F |uⁿ = 1}. And let n⁰ be the maximal order of elements in S. Clearly, n⁰ ≤ n It’s clear that S is an abelian multi- plicative group. Therefore, it’s easy to see that order of elements in S divides n⁰. It follows that uⁿ⁰ = 1 for all u ∈ S. Hence |S| ≤ n⁰.

Since we assume that (n, char(K)) = 1, therefore xⁿ− 1 is separable.

It follows that roots of xⁿ− 1 are all distinct, hence |S| = n. One sees that n = n⁰, therefore, there are elements of order n in S, denoted ζ.

It follows that F = K(S) = K(ζ).

For any σ ∈ Gal_F/K, σ(ζ) ∈ S. Hence σ(ζ) = ζⁱ for some i. There- fore, we have a natural map φ : Gal_F/K → Z_n by φ(σ) = i if σ(ζ) = ζⁱ. Note that if ζⁱ is not a primitive n-th root of unity, then K(ζⁱ) is not the splitting field of xⁿ− 1, hence not equal to K(ζ), which is absurd.

Thus sigma we conclude that ζⁱ is a primitive n-th root of unity. It’s easy to see that this is equivalent to (i, n) = 1. Thus φ : Gal_F/K → Z^∗_n is an injective group homomorphism.

Lastly, if n = p^k with p 6= 2 or if n = 2, 4, then Z^∗_n is cyclic. Hence

every subgroup is cyclic. ¤

The structure of cyclotomic extension is thus determined by the primitive n-th root of unity. It’s then natural to ask the degree of such extension and their minimal polynomials.

Definition 3.9.4. If charK - n, then the n-th cyclotomic polyno- mial over K is defined as:

gn(x) := Y

ζi: prim. n-th root of 1

(x − ζi).

Proposition 3.9.5. We have the following:

1. xⁿ− 1 =Q

d|ng_d(x).

2. g_n(X) ∈ P [x], where P denoted the prime field. Moreover, if charK = 0, we identify P = Q, then g_n(x) ∈ Z[x].

3. deg(g_n(x)) = ϕ(n), where ϕ denotes the Euler φ-function.

Proof. (3) is clear from the definition.

For (1), we consider the following decomposition of sets {ζⁱ}i=0,...,n−1= ∪_d|n{ζⁱ|o(ζⁱ) = d}.

Note that o(ζⁱ) = d implies that ζⁱ is a primitive d-th root of unity.

Thus we define g_d⁰(x) := Q

o(ζⁱ)=d(x − ζⁱ), and then g_d⁰(x)|g_d(x). By the decomposition, we have

xⁿ− 1 = Y

i=0,...,n−1

(x − ζⁱ) =Y

d|n

g⁰_d(x).

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Computing degrees, we have n =X

d|n

deg(g_d⁰(x)) ≤ X

d|n

deg(gd(x)) =X

d|n

ϕ(d) = n.

Therefore, g_d⁰(x) = g_d(x).

To see (2), we prove by induction on n. We assume that g_d(x) ∈ P [x]

for all d < n. We can write xⁿ− 1 = g_n(x)f (x) ∈ F [x]. In P [x], we have xⁿ−1 = f (x)q(x)+r(x) by the division algorithm. We shall prove that r(x) = 0 and thus g_n(x) = q(x) ∈ P [x] by the unique factorization of F [x].

It suffices to show that r(x) = 0. To this end, note that f (x)|xⁿ− 1 in F [x], and thus f (x)|r(x) in F [x]. However, deg(r(x)) < deg(f (x)) unless r(x) = 0. This completes the proof of (2).

When char(K) = 0, similar inductive argument plus Gauss Lemma

will work. We leave it to the readers. ¤

Finally, if K = Q then the cyclotomic extension behave even nicer.

Proposition 3.9.6. F = Q(ζ) be the n-th cyclotomic extension over Q. Then

1. g_n(x) is irreducible.

2. [F : bQ] = ϕ(n).

3. Gal_QF ∼= Z^∗_n.

Example 3.9.7.

Consider the 3-rd cyclotomic extension over F₇. Then g₃(x) = ^x_x−1³⁻¹ =

(x − 2)(x − 4) is not irreducible. ¤

Proof. Asuuming (1), then F = Q[ζ] is generated by ζ, where minimal polynomial of ζ over Q is g_n(x). Thus [Q[ζ] : Q] = deg(g_n(x)) = ϕ(n). Morover, for every i ∈ Z^∗_n, the map ζ 7→ ζⁱ produces an Q- automorphism of F . Thus (3) follows.

It thus suffices to prove (1). Recall that g_n(x) ∈ Z[x]. If g_n(x) = f (x)h(x) ∈ Z[x], where f (x) is an irreducible polynomial with f (ζ) = 0.

We claim that ζ^p is also a root of f (x) for all (p, n) = 1. Grant this claim, then by this process, we can conclude that ζⁱ is a root of f (x) for all (i, n) = 1. Therefore, f (x) = g_n(x) is irreducible.

We now prove the claim. Suppose on the contrary that ζ^p is not a root of f (x). Then it’s a root of h(x). We have h(ζ^p) = 0. Hence ζ is a root of h(x^p). Since f (x) is irreducible, it’s minimal polynomial of ζ over Q. We have f (x)|h(x^p). Thus we can write h(x^p) = f (x)k(x) for some k(x) in Q[x]. By Gauss’ Lemma, this equation holds in fact in Z[x]. We now consider ring homomorphism ¯:Z[x] → Z_p[x]. Then

(h(x))^p = h(x^p) = f (x)k(x).

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Thus g.c.d(h(x), h(x)) 6= 1 in Zp[x]. It follows that xⁿ− 1 = (xⁿ− 1

g_n(x) )f (x)h(x)

has multiple roots. But xⁿ− 1⁰ = n¯xⁿ⁻¹ 6= 0. So this is the required

contradiction. ¤