# 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 ∎

# 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.
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