Mathematics reveals its secrets only to those who approach it with pure love, for its own beauty, Archimedes
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]
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) = a_{0} + a_{1}x + ··· + a_{n}x^{n} → Φ(a_{0}) + Φ(a_{1})x + ··· + Φ(a_{n})x^{n}.
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 = {α_{1}, α_{2}, ···, α_{r}} where α_{i} = τ(α) for some τ ∈ Gal(K/F) and {σ_{k}(α_{1}), σ_{k}(α_{2}), ···, 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_{k}σ_{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.
τ(f) = [τ is an homomorphism] τ(x -α_{1})τ(x -α_{2})···τ(x -α_{r}) = [τ(x) = x] (x -τ(α_{1}))(x -τ(α_{2}))···(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].
g(α) = 0 ⇒ τ(g(α)) = 0 ⇒ [g ∈ F[x], say g = a_{0} +a_{1}x + ··· + a_{m}x^{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) = {σ_{1}, σ_{2},···, 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 = {α_{1}, α_{2}, ···, α_{r}} where S is the set of distinct elements among σ_{1}(α), σ_{2}(α),···, σ_{n}(α). The images of α under the F-automorphisms σ_{1}, σ_{2},···, σ_{n} need not all be distinct, so it is possible that r < n. α is among these elements α_{1}, α_{2}, ···, 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, σ_{1}, σ_{2}, σ_{3}} ≡ 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, α_{1}, α_{2}, ···, 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(β_{1}, β_{2}, ···, β_{n}), and let f_{i} be the irreducible polynomial of β_{i} over F, and consider f = f_{1}f_{2}···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 β_{1} over F is f_{1} = (x- β_{11})(x- β_{12})···(x- β_{1r1}), β_{1l} = β_{1} for some l, each β_{1i} = τ(β_{1}) for some τ ∈ Gal(K/F), τ: K → K is an automorphism, and in particular, τ(β_{1}) ∈ K, so every root of the irreducible polynomial of β_{1} over F is in K ⇒ f_{1} 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(β_{1}, β_{2}, ···, 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 = f_{1}f_{2}··· 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(f_{1}) = r > 1 without losing any generality.
Let α_{1} ∈ K\F be a root of f_{1} (f and f_{1} are both irreducible in F[x], i.e., they have no roots in F). We are interested in the intermediate field F(α_{1}). α_{1} ∉ F ⇒ #number of roots of f in K\F(α_{1}) < # number of roots of f in K\F (α_{1} is one of these)
F ⊆ F(α_{1}) ⊆ K, futhermore f_{1} is irreducible in F[x], deg(f_{1}) = r > 1 ⇒ f_{1} has no roots in F, α_{1} ∉ F
By the induction hypothesis, we conclude K/F(α_{1}) is Galois and therefore K^{Gal(K/F(α1))} = F(α_{1})
f is separable ⇒ f_{1} is separable, deg(f_{1}) = r > 1 ⇒ f_{1} has r distinct roots in K, say α_{1}, α_{2}, ···, 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(α_{1})≋F[x]/⟨f_{1}⟩≋F(α_{i}), σ_{i}: F(α_{1}) → F(α_{i}), defined by σ_{i}(α_{1}) = 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}(α_{1}) = 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(α_{1})) -this is trivial every automorphism of K that fixes F(α_{1}), 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(α_{1})
β ∈ F(α_{1}) ⇒ [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-1}a^{n-1} + c_{n-2}a^{n-2} + ··· + c_{1}a + c_{0}, where c_{0}, c_{1}, ···, c_{n-1} ∈ F ] An F-basis of F(α_{1}) is 1, α_{1}, α_{1}^{2}, ···, α_{1}^{r-1}
Therefore, β can be written uniquely as, β = a_{r-1}α_{1}^{r-1} + a_{r-2}α_{1}^{r-2} + ··· + a_{1}α_{1} + a_{0}, a_{i} ∈ F. Let’s apply σ_{1},···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-1}σ_{i}(α_{1})^{r-1} + a_{r-2}σ_{i}(α_{1})^{r-2} + ··· + a_{1}σ_{i}(α_{1}) + a_{0} ⇒ ∀i, 1 ≤ i ≤ r, β = a_{r-1}σ_{i}(α_{1})^{r-1} + a_{r-2}σ_{i}(α_{1})^{r-2} + ··· + a_{1}σ_{i}(α_{1}) + a_{0}
Let h(x) = a_{r-1}x^{r-1} + a_{r-2}x^{r-2} + ··· + a_{1}x + a_{0} -β ∈ F(α_{1})[x], deg(h(x)) ≤ r -1. But α_{1}, α_{2}, ···, α_{r} are all distinct (Recall: f_{1} has r distinct roots in K, say α_{1}, α_{2}, ···, α_{r}) roots of h(x) in K (α_{i} = σ_{i}(α_{1})) ⊥ (h(x) has at least r distinct roots) ⇒ h(x) = 0 ⇒ All its coefficients are zero, and in particular a_{0} -β = 0 ⇒ β = a_{0} ∈ F ⇒ β ∈ F ∎
Theorem. Let K/F be a finite extension. The following statements are equivalent:
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 σ_{1}, ···, σ_{n}: K → K be distinct field automorphisms. Suppose that σ_{1}, ···, σ_{n} forms a group under composition. If F is a fixed field of σ_{1}, ···, 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(α_{1},···, 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) = {σ_{1},···, 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 α_{1}, α_{2}, ···, 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 ∎