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/K^G).

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 q_E(x) over F. Φ can be naturally “extended”, Φ: E[x] → F[x], p(x) = a₀ + a₁x + ··· + a_nxⁿ → Φ(a₀) + Φ(a₁)x + ··· + Φ(a_n)xⁿ.

Let q_F(x) denotes the polynomial obtain by applying Φ and L be the splitting of q_F(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 p_E(x) be any irreducible factor of q_E(x), deg(p_E(x))>1, and having a root, say α (α ∈ K\F because K is the splitting field of the polynomial q_E(x)). Let call p_F(x) the polynomial obtained by applying the natural extension of Φ to the coefficients of p_E(x).

If ψ is any isomorphism from K to L, then ψ(α) is some root β_i of the polynomial p_F(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(β_i), $\barϕ_i|_E=Φ$ is equal to the number of roots β_i of p_F(x), which is at most [F(β) : F] = deg(p_F) = deg(p_E) = [E(α) : E], and equality is reached exactly when p_E 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(β_i) because K is the splitting field (q_E) over E(α), L is the splitting field (q_F) over F(β_i), 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 q_E(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 β_i of p_F(x) (that is, #number of $\barϕ_i$: E(α)→ F(β_i), $\barϕ_i|_E=Φ$) · [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 q_E(x) is separable ( ⇒ p_E(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, K^H ⊇ K^G.

Proof.

H ⊆ G ⇒ Let α ∈ K^G ⇒ σ(α) = α, ∀σ ∈ G ⇒ σ(α) = α ∀σ ∈ H ⇒ α ∈ K^H ⇒ K^G ⊆ K^H∎

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, ···, α_r} where α_i = τ(α) for some τ ∈ Gal(K/F) and {σ_k(α₁), σ_k(α₂), ···, σ_k(α_r)}, σ_k ∈ G = Gal(K/F) are identical or, alternatively, the function σ_k: S → S, α_i → σ_k(α_i) is bijective.

• Does it map to S? Let α_i ∈ S ⇒ ∃j: α_i = σ_j(α) ⇒ σ_k(α_i) = σ_k(σ_j(α)) = [G is group] (σ_kσ_j)(α) = σ_k’(α) ∈ S, that is, σ_k(α_i) is an element of S.

• σ_k is injective. If σ_k(α_i) = σ_k(α_j) ⇒ [σ_k ∈ Gal(K/F), automorphism] α_i = α_j. σ_k is injective ⇒ [S is a finite set] σ_k 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 -α_r) = [τ(x) = x] (x -τ(α₁))(x -τ(α₂))···(x -τ(α_r)) ⇒ [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), K^G = [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 -α_r) 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 + ··· + a_mx^m and τ fixes all coefficients of g] g(τ(α)) = 0 ∀τ ∈ Gal(K/F) ⇒ g(α_i)=0 ∀i = 1,···,r ⇒ [α_i 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,···, σ_n} ⇒ [Proposition 1, Theorem 2] F = K^Gal(K/F) = K^G. [K : F] = |G| = n.

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

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, ···, α_r (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, ···, β_n), and let f_i be the irreducible polynomial of β_i over F, and consider f = f₁f₂···f_n. By the previous analysis, ∀i f_i 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- β_1r1), β_1l = β₁ for some l, each β_1i = 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 f_i 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, ···, β_n)) ⇒ 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 = K^Gal(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··· f_s ∈ F[x] be the irreducible factorization of f in F[x]. Since n> 0, at least one f_i is such that deg(f_i) > 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 K^{Gal(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, ···, α_r.

[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(α_i), σ_i: F(α₁) → F(α_i), defined by σ_i(α₁) = α_i 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 σ_i: K → K automorphism such that σ_i|_F = identity (fixes F) and σ_i(α₁) = α_i, so σ_i ∈ Gal(K/F) ∀i = 1, ··· r.

Claim: K/F is Galois.

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

β ∈ K^Gal(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] K^Gal(K/F) ⊆ K^{Gal(K/F(α₁))}

β ∈ K^Gal(K/F) ⊆ K^{Gal(K/F(α₁))} ⇒ β ∈ K^{Gal(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, ···, a^n-1}, i.e., every member of F(a) can be written in the form c_n-1a^n-1 + c_n-2a^n-2 + ··· + c₁a + c₀, where c₀, c₁, ···, c_n-1 ∈ F ] An F-basis of F(α₁) is 1, α₁, α₁², ···, α₁^r-1

Therefore, β can be written uniquely as, β = a_r-1α₁^r-1 + a_r-2α₁^r-2 + ··· + a₁α₁ + a₀, a_i ∈ F. Let’s apply σ₁,···σ_r to β:

[σ_i ∈ Gal(K/F), β ∈ K^Gal(K/F) ⊆ K^{Gal(K/F(α₁))}] σ_i(β) = β. Besides, σ_i(β) = a_r-1σ_i(α₁)^r-1 + a_r-2σ_i(α₁)^r-2 + ··· + a₁σ_i(α₁) + a₀ ⇒ ∀i, 1 ≤ i ≤ r, β = a_r-1σ_i(α₁)^r-1 + a_r-2σ_i(α₁)^r-2 + ··· + a₁σ_i(α₁) + a₀

Let h(x) = a_r-1x^r-1 + a_r-2x^r-2 + ··· + a₁x + a₀ -β ∈ F(α₁)[x], deg(h(x)) ≤ r -1. But α₁, α₂, ···, α_r are all distinct (Recall: f₁ has r distinct roots in K, say α₁, α₂, ···, α_r) roots of h(x) in K (α_i = σ_i(α₁)) ⊥ (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 = K^Gal(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 = K^Gal(K/F) ⇒ [Degree of Fixed Fields. Let K be any field, and let σ₁, ···, σ_n: K → K be distinct field automorphisms. Suppose that σ₁, ···, σ_n forms a group under composition. If F is a fixed field of σ₁, ···, σ_n. 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,···, α_n) for some α_i algebraic over F. Let f_i(x) be the irreducible polynomial of α_i ∀i ⇒ [K/F normal, α_i ∈ K by construction] f_i splits completely in K and [K/F separable] all roots are distinct. Let f(x) be the least common multiple of the f_i ⇒ 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 = K^Gal(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,···, σ_n})} 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 K^Gal(K/F) =[K/F Galois] F ⇒ f ∈ F[x]

f is a monic polynomial in F[x], such that f(α) = 0 (where α_i = 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, ···, α_r ⇒ 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 ∎

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
5. Fields and Galois Theory. Morandi. P., Springer.
6. Fields and Galois Theory. By Evan Dummit, 2020.