Advanced Algebra I
Galois groups of polynomials, cyclotomic extension We first recall something about separable extension. The main pur- pose is to prove the following proposition we used in the previous sec- tion.
Proposition 0.1. Suppose that F = K(S) such that each elements of S is separable over K, then F/K is separable.
To start with, let f (x) be an irreducible polynomial in K[x] and f0(x) be its derivative (formally). More precisely, if f (x) = sumni=0aixi, then f0(x) :=Pn
i=1iaixi−1. One has the following equivalence:
(1) f (x) is separable, i.e. no multiple roots in K.
(2) (f (x), f0(x)) = 1 ∈ K[x].
(3) (f (x), f0(x)) = 1 ∈ K[x].
(4) f0(x) = 0.
Therefore, the only possibility to have non-separable polynomial is char(K) = p and f (x) = g(xp).
Given an element u algebraic over K, one can define the separable degree to be the number of distinct roots of minimal polynomial. This notion can be extended to a general setting:
Definition 0.2. Let F/K be an extension. Fix an embedding σ : K → L = L. We define the separable degree of F/K, denoted [F : K]s, to be the cardinality of
Sσ := {τ : F → L|τ|K = σ}.
One can check that [F : K]s is independent of σ and L. Hence the definition is well-defined. Moreover, if F = K(u) for u algebraic over K, then [F : K]s = [K(u) : K]s is the number of distinct roots of the minimal polynomial p(x) of u. (By considering K-embedding τ : K(u) → K, τ (u) must be a root of p(x) and τ is determined by τ (u)).
Proposition 0.3. If K ⊂ E ⊂ F , then [F : K]s = [F : E]s[E : K]s. Moreover, if F/K is finite, then [F : K]s ≤ [F : K].
Proof. Fix an embedding σ : K → L, there are extensions {σi}i∈I : E → L with |I| = [E : K]s. And for a fix σi, there are extensions {σi,j}j∈J : F → L with |J| = [F : K]s. Thus
[F : K]s = |I| · |J| = [F : E]s[E : K]s.
If F/K is finite, then F = K(u1, ..., ut). It’s clear that for all i, [K(u1, ..., ui) : K(u1, ..., ui−1)]s ≤ [K(u1, ..., ui) : K(u1, ..., ui−1)] since the number of distinct root is less or equal than degree of minimal
polynomial. ¤
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Then we have the following useful criterion:
Proposition 0.4. If F/K is finite, then F/K is separable if and only if [F : K]s = [F : K].
Proof. Write F = K(u1, ..., ut). If F/K is separable, then for all i, [K(u1, ..., ui) is separable over K(u1, ..., ui−1)]. Then we have for all i
[K(u1, ..., ui) : K(u1, ..., ui−1]s = [K(u1, ..., ui) : K(u1, ..., ui−1].
It follows that [F : K]s= [F : K].
On the other hand, for any u ∈ F , we have
[F : K]s= [F : K(u)]s[K(u) : K]s ≤ [F : K(u)][K(u) : K] = [F : K].
Hence all the above are in fact equality, therefore [K(u) : K]s= [K(u) :
K] and u is separable over K. ¤
Proof of the Proposition 1. For any u ∈ K(S), we may assume that u ∈ K(u1, ..., ut) for some u1, ..., ut ∈ S separable over K. It’s obvious that K(u1, ..., ui) separable over K(u1, ..., ui−1) for all i. Thus one has that [K(u1, ..., ut) : K]s = [K(u1, ..., ut) : K] and thus K(u1, ..., ut) is
separable over K. Thus so is u. ¤
We now study the Galois groups of polynomials. First of all, for a given f (x) ∈ K[x], we define the Galois group of f (x) over K, denoted Gf, to be the Galois group of a splitting field F over K.
Theorem 0.5. (1) If deg(f (x)) = n, then Gf ,→ Sn.
(2) If f (x) is irreducible and separable of degree n, then Gf is tran- sitive in Sn and n¯
¯|Gf|.
Proof. For every σ ∈ Gf, σ induces a permutation on roots of f (x).
Hence the mapping by sending σ to the corresponding permutation gives the required embedding.
Let ui, uj be two distinct roots of f (x), then we have an isomorphism σ : K(ui) → K(uj) such that σ(ui) = uj. Since the splitting field F is normal, thus σ can be extended to an automorphism of F . Therefore, we have find a K-automorphism sending ui to uj for any i, j. Hence it’s transitive in Sn.
Moreover, one sees that [Gf : K(ui)0] = [K(ui) : K] = n. Thus n¯
¯|Gf|. ¤
Example 0.6. The only transitive subgroups in S3 are S3 and A3. The only transitive subgroups of order divisible by 4 in S4 are S4, A4, ∼= D4, V, ∼= Z4.
For a given polynomial f (x) with roots u1, ..., un. one can define
∆ := Y
i<j
(ui− uj).
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Then ∆ is preserved by even permutations, i.e. Gf∩An. Then D := ∆2 is preserved by all Gf. One can see that D is in K. D is called the discriminant of f (x). Assume that f is irreducible and separable, then F is Galois over K. One can see that (Gf ∩ An)0 = K(∆) if char(K) 6= 2. Applying this to degree 3, we have:
Theorem 0.7. Let f (x) be an irreducible separable polynomial of de- gree 3 over K with char(K) 6= 2 ,then Gf = A3 if and only if D is a perfect square in K and Gf = S3 if and only if D is not a perfect square.
We also remark that D is computable. For example, if f (x) = x3+ px + q, then D = −4p2− 27q3.
The story for degree 4 are similar but more delicate. Let f (x) be an irreducible separable polynomial of degree 4. Let u1, ..., u4 be the roots of f (x). And let F = K(u1, ..., u4) be the splitting field, which is Galois over K. Let α := u1u2+ u3u4, β := u1u3+ u2u4, γ := u1u4+ u2u3. It’s clear that they are all distinct since f (x) is separable. We can consider an intermediate field E := K(α, β, γ). Let g(x) := (x−α)(x−β)(x−γ).
One sees that g(x) ∈ K[x] and E is s splitting field of g(x). One can check directly that E0 = Gf ∩ V . Thus we have:
Theorem 0.8. Keep the notation as above. Let m := [E : K]. Then (1) m = 6 ⇔ Gf = S4.
(2) m = 3 ⇔ Gf = A4. (3) m = 1 ⇔ Gf = V .
(4) m = 2, f (x) is irreducible in E[x] ⇔ Gf ∼= D4. (5) m = 2, f (x) is reducible in E[x] ⇔ Gf ∼= Z4.
Proof. As we have seen that the only transitive subgroup with order di- visible by 4 are S4, A4, V4, ∼= D4, ∼= Z4. Hence the first three equivalence are trivial.
If m = 2 then Gf ∼= D4 or Gf ∼= Z4. If f (x) is irreducible in E[x], then f (x) is the minimal polynomial of u1. Hence [F : E] = [F : E(u1)][E(u1) : E] = [F : E(u1)]4. In particular, 4¯
¯[F : E] = |Gf ∩ V |.
Therefore, Gf ∼= D4.
On the other hand, if f (x) is reducible, then f (x) can not have factor of degree 3 since 3 6¯
¯|Gf ∩ V |. ¤
We now start the study of cyclotomic extension.
Definition 0.9. A cyclotomic extension of order n over K is a splitting field of xn− 1.
Remark 0.10. If char(K) = p and n = prm, then xn−1 = (xm−1)pr. Hence we may assume that either char(K) = 0 or char(K) - p in the study of cyclotomic extension.
The main theorem is the following
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Theorem 0.11. 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 GalF/K can be identified as a subgroup of Z∗n.
(3) If n is prime, then GalF/K is cyclic.
Proof. Let S := {u ∈ F |un = 1}. And let n0 be the maximal order of elements in S. It’s clear that S is an abelian multiplicative group.
Therefore, it’s easy to see that order of elements in S divides n0. It follows that un0 = 1 for all u ∈ S.
Since we assume that (n, chat(K)) = 1, therefore xn− 1 is separable,
|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 ∈ GalF/K, σ(ζ) ∈ S. Hence σζ = ζi for some i. Therefore, we have a natural map φ : GalF/K → Zn by φ(σ) = i if σ(ζ) = ζi. Since σ are automorphism, one has image in Z∗n. It’s easy to see that φ : GalF/K → Z∗n is an injective group homomorphism.
Lastly, if n is prime, then Z∗p is cyclic. Hence every subgroup is
cyclic. ¤
