    # Characterization of Galois extensions

Mathematics reveals its secrets only to those who approach it with pure love, for its own beauty, Archimedes

# Recall

Proposition 1. Let K be any field and let G be a finite group of automorphisms of K. Suppose F is the fixed field of G. Then G = Gal(K/F) = Gal(K/KG).

Theorem 2. A finite extension K/F is Galois iff |Gal(K/F)| = [K : F]

# Characterization of Galois extensions

Automorphisms of Splitting Fields. If K is a splitting field over F, then |Gal(K/F)| ≤ [K : F] with equality if and only if K/F is separable, i.e., when K is the splitting field of a separable polynomial over F.

Proof. (Based on Fields and Galois Theory. By Evan Dummit, 2020.)

Let’s induct on n = [K : F]

Suppose Φ: E → F is a field isomorphism and K is the splitting field of the polynomial qE(x) over F. Φ can be naturally “extended”, Φ: E[x] → F[x], p(x) = a0 + a1x + ··· + anxn → Φ(a0) + Φ(a1)x + ··· + Φ(an)xn.

Let qF(x) denotes the polynomial obtain by applying Φ and L be the splitting of qF(x) over F. By [Extension Theorem. Let ϕ: E→ F be an isomorphism of fields and let p(x) be a non-constant polynomial in E[x] and q(x) the corresponding polynomial in F[x] under ϕ. If K and L are splitting field of p(x) and q(x) respectively, then ϕ extends to an isomorphism ψ: K → L, i.e., ψ|E = Φ ] we can extend Φ to an isomorphism ψ: K → L. We claim that the number of such isomorphisms ψ: K → L is at most [K : F] with equality if and only if K/F is separable. The result will follow upon setting Φ as the identity map and E = F.

Case base, n = 1 = [K : F] ⇒ K = E, L = F ⇒ The only possible map ψ: K → L extending Φ is Φ itself, just 1.

Inductive case. Suppose n ≥ 2, let pE(x) be any irreducible factor of qE(x), deg(pE(x))>1, and having a root, say α (α ∈ K\F because K is the splitting field of the polynomial qE(x)). Let call pF(x) the polynomial obtained by applying the natural extension of Φ to the coefficients of pE(x).

If ψ is any isomorphism from K to L, then ψ(α) is some root βi of the polynomial pF(x), which is obviously in L. By [Extension theorems Let ϕ: E → F be an isomorphism of fields. Let K be an extension field of E and α ∈ K be algebraic over E with minimal polynomial p(x). Suppose that L is an extension field of F such that β(∈L) is root of Φ(p(x)) in F[x]. Then, ϕ extends to a unique isomorphism $\barϕ :E(α)→F(β)$ such that $\barϕ(α)=β~ and~ \barϕ$ agrees with Φ on E], the number of isomorphisms $\barϕ_i$: E(α)→ F(βi), $\barϕ_i|_E=Φ$ is equal to the number of roots βi of pF(x), which is at most [F(β) : F] = deg(pF) = deg(pE) = [E(α) : E], and equality is reached exactly when pE is separable (all its roots are distinct).

Futhermore, we can apply the inductive hypothesis to each of the possible maps $\barϕ_i$: E(α)→ F(βi) because K is the splitting field (qE) over E(α), L is the splitting field (qF) over F(βi), and [E(α): F] < [K : F] because α ∈ K\F and E(α) is an intermediate field, and realize that the number of isomorphisms ψ: K → L extending $\barϕ_i$ is at most [K : E(α)], and equality is achieved precisely when qE(x) is separable.

If we consider all the maps $\barϕ_i$, the total number of isomorphisms ψ: K → L extending Φ: E → F is at most #number of roots βi of pF(x) (that is, #number of $\barϕ_i$: E(α)→ F(βi), $\barϕ_i|_E=Φ$) · [K : E(α)] (that is, #number of isomorphisms ψ: K → L extending $\barϕ_i$) = [E(α) : E]·[K : E(α)] = [K : E] and equality is achieved if and only if qE(x) is separable ( ⇒ pE(x) is separable, too)∎

Lemma. Fixed field inclusion reversing. Let K be a field and H ⊆ G be two groups of automorphisms of K. Then, KH ⊇ KG.

Proof.

H ⊆ G ⇒ Let α ∈ KG ⇒ σ(α) = α, ∀σ ∈ G ⇒ σ(α) = α ∀σ ∈ H ⇒ α ∈ KH ⇒ KG ⊆ KH

Lemma. Given a finite and Galois extension K/F and some α ∈ K, the Galois group acts or permutes on the set S:= {σ(α): σ ∈ G}, that is, the orbit of α under the action of G (where given σ ∈ G, α ∈ K, σ·α = σ(α))

Proof.

The sets S = {α1, α2, ···, αr} where αi = τ(α) for some τ ∈ Gal(K/F) and {σk1), σk2), ···, σkr)}, σk ∈ G = Gal(K/F) are identical or, alternatively, the function σk: S → S, αi → σki) is bijective.

• Does it map to S? Let αi ∈ S ⇒ ∃j: αi = σj(α) ⇒ σki) = σkj(α)) = [G is group] (σkσj)(α) = σk’(α) ∈ S, that is, σki) is an element of S.

• σk is injective. If σki) = σkj) ⇒ [σk ∈ Gal(K/F), automorphism] αi = αj. σk is injective ⇒ [S is a finite set] σk is bijective ∎

Fundamental Lemma. Given a finite and Galois extension K/F and some α ∈ K, we can write the minimal polynomial of α over F as $\prod_{α_i∈S}(x-α_i)$ where S:= {σ(α): σ ∈ G}.

In other words, the minimal polynomial is defined to be the one whose roots are precisely all the distinct Galois conjugates of α.

Proof.

• f = $\prod_{i∈S}(x-α_i)$ ∈ F[x]. Let us apply τ to f for every τ ∈ G. τ: K → K induces in a natural way, τ: K[x] → K[x] homomorphism, τ(x) = x.

τ(f) = [τ is an homomorphism] τ(x -α1)τ(x -α2)···τ(x -αr) = [τ(x) = x] (x -τ(α1))(x -τ(α2))···(x -τ(αr)) ⇒ [Since τ is a bijection in S -Lemma-, τ permutes the values of S] = f. Therefore, ∀τ ∈ G, τ(f) = f, then every coefficient of f is fixed by every element of G, so every coefficient of f lie in the fixed field of Aut(K/F), KG = [K/F is Galois if the fixed field by the automorphism (Galois) group Aut(K/F) is precisely the base field F] F ⇒ f ∈ F[x].

• f is the irreducible polynomial of α over F. Let g ∈ F[x] be the irreducible polynomial of α over F ⇒ [f = (x -α1)(x - α2)···(x -αr) and α ∈ S, α is a root of f] f is divisible by the irreducible (minimal) polynomial.

g(α) = 0 ⇒ τ(g(α)) = 0 ⇒ [g ∈ F[x], say g = a0 +a1x + ··· + amxm and τ fixes all coefficients of g] g(τ(α)) = 0 ∀τ ∈ Gal(K/F) ⇒ g(αi)=0 ∀i = 1,···,r ⇒ [αi are distinct elements] g has at least r roots ⇒ deg(g) ≥ r.

deg(f) = r, g divides f ⇒ deg(g) ≤ r ⇒ [We know that deg(g) ≥ r] deg(g) = r ⇒ g = f ⇒ f is the irreducible polynomial of α over F.

Theorem. Let K/F be a finite field extension. Then, K/F is Galois if and only if K is the splitting field of a separable polynomial over F.

Remark: K needs to be the splitting field of a separable polynomial over F, e.g., F = $\mathbb{F_2(t)}$ is normal, but not Galois.

Proof.

⇒) Suppose K/F is Galois, let G = Gal(K/F) = {σ1, σ2,···, σn} ⇒ [Proposition 1, Theorem 2] F = KGal(K/F) = KG. [K : F] = |G| = n.

Let α ∈ K be any arbitrary element. Consider S = {α1, α2, ···, αr} where S is the set of distinct elements among σ1(α), σ2(α),···, σn(α). The images of α under the F-automorphisms σ1, σ2,···, σn need not all be distinct, so it is possible that r < n. α is among these elements α1, α2, ···, αr since the identity map is indeed an automorphism in Gal(K/F).

We can consider the action of Aut(K/F) = Gal(K/F) on K given by τ·α = τ(α), and S is the orbit of α under this action.

Example, let K = $ℚ(\sqrt{2},i), F = ℚ$, G = {1, σ1, σ2, σ3} ≡ ℤ/2ℤ x ℤ/2ℤ. Let $α = \sqrt{2},$ S ={$\sqrt{2}, -\sqrt{2}$}, r = 2 < 4 = n

Lemma. Let f = $\prod_{α_i∈S}(x-α_i)$ where S is the set defined as S:= {σ(α): σ ∈ G}, then f is the irreducible polynomial of α over F. Futhermore, α is separable over f.

In our working example, f = $(x-\sqrt{2})(x+\sqrt{2}) = x^2-2$. This is a useful way to find the irreducible polynomial of an element in a Galois extension.

By the Fundamental Lemma, f is the irreducible polynomial of α over F.

Claim: α is separable over F. Therefore, f (deg(f) = r) has r distinct roots in K, namely, α1, α2, ···, αr (α is one of them since the identity is an automorphism) ⇒ f is separable ⇒ α is separable over F∎

Because we have taken an arbitrary α ∈ K, we have demonstrated that K/F is separable.

K/F is a finite extension. Let K = F(β1, β2, ···, βn), and let fi be the irreducible polynomial of βi over F, and consider f = f1f2···fn. By the previous analysis, ∀i fi is separable, and so is f.

Let’s prove that K is the splitting field of f over F. We know that the irreducible polynomial of β1 over F is f1 = (x- β11)(x- β12)···(x- β1r1), β1l = β1 for some l, each β1i = τ(β1) for some τ ∈ Gal(K/F), τ: K → K is an automorphism, and in particular, τ(β1) ∈ K, so every root of the irreducible polynomial of β1 over F is in K ⇒ f1 splits completely over K. Similarly fi splits completely over K for each i ⇒ f splits completely over K, K is generated by the roots of f over F (K = F(β1, β2, ···, βn)) ⇒ K is the splitting field of f over F, and we have already shown that f is separable∎

Remark: We have also demonstrated that a Galois extension is both normal and separable.

⇐) There are two options: see above the Automorphisms of Splitting Fields theorem, based on Fields and Galois Theory by Evan Dummit, or the following proof by NPTEL-NOC IITM, Introduction to Galois Theory.

Suppose that K is the splitting field of a separable polynomial f ∈ F[x] over F. We claim that K/F is Galois. [K/F is Galois if the fixed field by the automorphism (Galois) group Aut(K/F) is precisely the base field F] F = KGal(K/F)

f splits completely in K[x]. Let’s induct on n = #number of roots of f in K\F.

Case base: n = 0. Then, there are no roots of f in K\F ⇒ all roots of f are in F ⇒ f splits completely in F[x] ⇒ [Since K is the splitting field of f] K = F, and trivially the extension F/F is Galois of degree 1.

Let’s assume that n > 0. The induction hypothesis is as follows: Given any intermediate field L (F ⊆ L ⊆ K) such that f (f ∈ F[x] ⊆ L[x]) has fewer than n roots in K\L, then K/L is Galois.

K is the splitting field of a separable polynomial f ∈ F[x] ⊆ L[x], so K is also the splitting field of f over L, and we are going to assume, as our induction hypothesis, that if the number of roots of f that are in K and not in L are less than n, then the extension K/L is Galois.

Let f = f1f2··· fs ∈ F[x] be the irreducible factorization of f in F[x]. Since n> 0, at least one fi is such that deg(fi) > 1 (otherwise, f is the product of linear polynomials and splits completely in F, but n = #number of roots of f in K\F > 0 ⊥), and let’s assume deg(f1) = r > 1 without losing any generality.

Let α1 ∈ K\F be a root of f1 (f and f1 are both irreducible in F[x], i.e., they have no roots in F). We are interested in the intermediate field F(α1). α1 ∉ F ⇒ #number of roots of f in K\F(α1) < # number of roots of f in K\F (α1 is one of these)

F ⊆ F(α1) ⊆ K, futhermore f1 is irreducible in F[x], deg(f1) = r > 1 ⇒ f1 has no roots in F, α1 ∉ F

By the induction hypothesis, we conclude K/F(α1) is Galois and therefore KGal(K/F(α1)) = F(α1)

f is separable ⇒ f1 is separable, deg(f1) = r > 1 ⇒ f1 has r distinct roots in K, say α1, α2, ···, αr.

[Theorem. Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial over F, deg(p(x)) = n. If a is a root of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/⟨p(x)⟩, F(a)≈F[x]/⟨p(x)⟩.] We can construct some automorphisms of Gal(K/F) as follows, ∀i=1,···, r, F(α1)≋F[x]/⟨f1⟩≋F(αi), σi: F(α1) → F(αi), defined by σi1) = αi and we [Extension Theorem. Let ϕ: E→ F be an isomorphism of fields and let p(x) be a non-constant polynomial in E[x] and q(x) the corresponding polynomial in F[x] under ϕ. If K and L are splitting field of p(x) and q(x) respectively, then ϕ extends to an isomorphism ψ: K → L ] can extend them to σi: K → K automorphism such that σi|F = identity (fixes F) and σi1) = αi, so σi ∈ Gal(K/F) ∀i = 1, ··· r.

Claim: K/F is Galois.

K/F is Galois ↭ KGal(K/F) = F. Trivially, F ⊆ KGal(K/F). It suffices to show that KGal(K/F) ⊆ F. Let β ∈ KGal(K/F), let’s prove that β ∈ F.

β ∈ KGal(K/F) ⇒ [By definition] every element of Gal(K/F) fixes β.

In our theorem, Gal(K/F) ⊇ Gal(K/F(α1)) -this is trivial every automorphism of K that fixes F(α1), fixes F, too- ⇒ [Lemma. Fixed field inclusion reversing] KGal(K/F) ⊆ KGal(K/F(α1))

β ∈ KGal(K/F) ⊆ KGal(K/F(α1)) ⇒ β ∈ KGal(K/F(α1)) = [By our induction hypothesis] F(α1)

β ∈ F(α1) ⇒ [Theorem. Let F be a field and let p(x) ∈ F[x] be an irreducible polynomial over F, deg(p(x)) = n. If a is a root of p(x) in some extension E of F, then F(a) is isomorphic to F[x]/⟨p(x)⟩, F(a)≈F[x]/⟨p(x)⟩. Moreover, F(a) is a vector space over F with a basis {1, a, ···, an-1}, i.e., every member of F(a) can be written in the form cn-1an-1 + cn-2an-2 + ··· + c1a + c0, where c0, c1, ···, cn-1 ∈ F ] An F-basis of F(α1) is 1, α1, α12, ···, α1r-1

Therefore, β can be written uniquely as, β = ar-1α1r-1 + ar-2α1r-2 + ··· + a1α1 + a0, ai ∈ F. Let’s apply σ1,···σr to β:

i ∈ Gal(K/F), β ∈ KGal(K/F) ⊆ KGal(K/F(α1))] σi(β) = β. Besides, σi(β) = ar-1σi1)r-1 + ar-2σi1)r-2 + ··· + a1σi1) + a0 ⇒ ∀i, 1 ≤ i ≤ r, β = ar-1σi1)r-1 + ar-2σi1)r-2 + ··· + a1σi1) + a0

Let h(x) = ar-1xr-1 + ar-2xr-2 + ··· + a1x + a0 -β ∈ F(α1)[x], deg(h(x)) ≤ r -1. But α1, α2, ···, αr are all distinct (Recall: f1 has r distinct roots in K, say α1, α2, ···, αr) roots of h(x) in K (αi = σi1)) ⊥ (h(x) has at least r distinct roots) ⇒ h(x) = 0 ⇒ All its coefficients are zero, and in particular a0 -β = 0 ⇒ β = a0 ∈ F ⇒ β ∈ F ∎

Theorem. Let K/F be a finite extension. The following statements are equivalent:

1. K/F is Galois.
2. F is the fixed field of Aut(K/F), i.e., F = KGal(K/F)
3. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., |Gal(K/F)| = [K : F]
4. K/F is a normal, finite, and separable extension.
5. K is the splitting field of a separable polynomial f ∈ F[x] over F.
6. Let Char(F)=0 or F be finite (or more generally F is perfect). Then, a finite extension K/F is Galois if and only if K/F is normal.

Proof.

6 ↭ 4) If Char(F) = 0, finite, or perfect, it is automatically separable.

3 ⇒ 2) Suppose |Gal(K/F)| = [K : F]. Suppose E is the fixed field of Gal(K/F), i.e., E = KGal(K/F) ⇒ [Degree of Fixed Fields. Let K be any field, and let σ1, ···, σn: K → K be distinct field automorphisms. Suppose that σ1, ···, σn forms a group under composition. If F is a fixed field of σ1, ···, σn. Then, [K: F] = n] |Gal(K/F)| = n = [K : E] = [because E is an intermediate field between K and F, [K:F]=[K:E]·[E:F]] [K : F]/[E : F] ⇒ [By assumption, |Gal(K/F)| = [K : F]] [E : F] = 1 ⇒ F is the fixed field of Gal(K/F)

4 ⇒ 5) K/F is a normal, finite, and separable extension. Then, [K/F finite extension] K = F(α1,···, αn) for some αi algebraic over F. Let fi(x) be the irreducible polynomial of αi ∀i ⇒ [K/F normal, αi ∈ K by construction] fi splits completely in K and [K/F separable] all roots are distinct. Let f(x) be the least common multiple of the fi ⇒ f(x) is separable, all of its roots are in K and generate K/F, so K is the splitting field of f in F

2 ⇒ 4) Suppose F = KGal(K/F). Let α ∈ K be an arbitrary element, f = $\prod_{α_i∈S}(x-α_i)$ ∈ K[x] where S:= {σ(α): σ ∈ Gal(K/F) = {σ1,···, σn})} is the irreducible polynomial of α over F as it has been previously demonstrated.

Recall: ∀τ ∈ Gal(K/F), τ permutes the element of S, hence if τ(f) denotes the result of applying τ to the coefficients of f, then τ(f) = $\prod_{α_i∈S}(x-τ(α_i))=f$ ⇒ The coefficients of f lie in the fixed field KGal(K/F) =[K/F Galois] F ⇒ f ∈ F[x]

f is a monic polynomial in F[x], such that f(α) = 0 (where αi = α for some i because identity is always an automorphism inside the Galois group), f splits completely in K[x] and f has r distinct roots in K, namely α1, α2, ···, αr ⇒ Since we can find such an f for any α ∈ K, it follows that K/F is both separable and normal∎

1 ⇐ 4) K/F is normal ⇒ K is the splitting field of a polynomial f over F. [K/F separable] Such a polynomial f is separable. In other words, K is the splitting field of a separable polynomial f over F ⇒ [Characterization of Galois extension theorem] K/F is Galois ∎

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