There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else — but persistent, Raoul Bott.

Recall. **Equivalence of Definition of Normal Extensions**. Let K/F be a finite extension of fields. Let $\bar \mathbb{F}$ be an algebraic closure of F that contains K (All algebraic closures of a field are isomorphic, so you can take *any arbitrary algebraic closure that contains K*). Then, the following conditions, any of which can be regarded as a definition of normal extension, are equivalent:

- Every embedding of K in $\bar \mathbb{F}$, σ: K → $\bar \mathbb{F}$, induces an F-automorphism from K to itself.
Recall: An embedding is a ring homomorphism σ: E → F. The Ker(σ) is an

**ideal of E**which cannot be the whole field E, because 1 ∉ Ker(σ) -σ(1) = 1-. The only ideas in fields are the zero or trivial ideal and the whole field itself ⇒ Ker(σ) = {0} ⇒ σ is an injective homomorphism, and E is isomorphic to the subfield σ(E) of F, E ≋ σ(E),**this justifies the name embedding**. - Every irreducible polynomial f ∈ F[x] with at least one root in K, splits (into a product of linear factors) completely in K[x].
- K is the splitting field of a polynomial f ∈ F[x].

Definition. If E is an extension field of F and if f(x) ∈ F[x], then **f splits over E if f(x) = a $\prod_i(x-α_i)$** for some α_{1},···, α_{n}∈ E, and a ∈ F, i.e., f factors completely into linear factors in K[x].

- Let f(x) ∈ F[x] have degree n. There is an extension field E of F with [E : F] ≤ n such that E contains a root of f. In addition, there is a field L containing F with [L : F] ≤ n! such that f splits over L. (Fields and Galois Theory. Morandi. P., Springer.)

Prop.

Let p(x) be an irreducible factor of f(x) in F[x], define E as the field F[x]/⟨p(x)⟩. Then F is isomorphic to a subfield of E. The map Φ: F → E = F[x]/⟨p(x)⟩ given by Φ(a) = a + ⟨p(x)⟩ is an injection of fields (F ≋ F[x]/⟨p(x)⟩). We will view F ⊆ E by replacing F with Φ(F). If α = x + ⟨p(x)⟩ ∈ K, then p(α) = p(x) + ⟨p(x)⟩ = [a+H = H iff a ∈ H] 0 + ⟨p(x)⟩, therefore α is a root of p in K, and a root of f. Besides, [K : F] = deg(p) ≤ n.

Let’s use induction on n.

n = 1, f = αx + β, with α, β ∈ F, so f already splits in F, and [F : F] = 1 ≤ 1!.

We already know that there is a field E ⊇ F with [E : F] ≤ n such that E contains a root, say α, of f(x) ⇒ f(x) = (x -α)g(x) with g(x) ∈ E[x], deg(g) < deg(f). By induction, there is a field L ⊇ E with [L : E] ≤ (n-1)! such that g splits over L, but then f splits over L, too, and [L : F] = [L : E][E : F] ≤ (n-1)!n = n!∎

If f(x) ∈ F[x], then E is a splitting field of f over F if f splits over E and E = F(α_{1},···, α_{n}), where α_{1},···, α_{n} are the roots of f. Futhermore, if f_{1}(x),…, f_{n}(x)∈ F[x], then there is a splitting field for {f_{1}(x),…, f_{n}(x)} over F, namely the splitting field of the product f = f_{1}(x)···f_{n}(x). Notice: We know there is a field E ⊇ F such that f splits over L, let α_{1}, ···,α_{n} be the roots of f. Then F(α_{1}, ···,α_{n}) is a splitting field for f over F.

- If K is a normal extension of F, and L is an intermediate extension (that is, F ⊂ L ⊂ K) then K is normal extension of L. However, L/F is not necessarily normal.

Proof.

Pretty trivial. K is the splitting field of f ∈ F[x] over F ⇒ f ∈ L[x] and K is the splitting field of f over L ⇒ K/L is normal∎

However, **L/F is not necessarily normal**, e.g., $\mathbb{Q}(\sqrt[4]{2}, i)/\mathbb{Q}$ is normal (it is the splitting field of x^{4}-2 over ℚ), but $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}$ is not normal.

Futhermore, the relation “is a normal extension of” is not transitive. Let K ⊆ L ⊆ F be finite extensions. **K/L and L/F normal does not imply that K/F is normal either**.

- If E and L are normal extensions of F, then E ∩ L is also a normal extensions of F.

Proof.

Take any irreducible polynomial in F[x], say m(x), that has a root in E ∩ L ⇒ By normality of E and L, such a polynomial will split into linear factors over both E and L, say m(x) = (x - e_{1})(x - e_{2})···(x - e_{d}) = (x - l_{1})(x - l_{2})···(x - l_{d}). Since each of these factorizations of f are in F[x], which is a UFD, these l_{i} are just a reordering of the m_{j}, and thus m(x) splits into linear factors in E ∩ L.

- If E and L are normal extensions of F contained in K, then the composite field of E and L, EL is also a normal extension.

Recall a composite or compositum of fields. Let K be a field, E and L be subfields of K. Then, the composite field of E and L is defined to be the intersections of all subfields of K containing both E and L.

Let σ be a homomorphism EL → $\bar \mathbb{F}$ that fixes F. Let’s apply σ on any arbitrary element $\sum_{i=1}^n a_ib_i$ ∈ EL where a_{i} ∈ E, b_{i} ∈ L (we are using that both E and L are normal and finite extensions of F, E, L ⊆ $\bar F$), $σ(\sum_{i=1}^n a_ib_i) = \sum_{i=1}^n σ(a_i)σ(b_i)$ ∈ EL because E, L are normal extensions of F(σ(E)⊂E, σ(L)⊂L). This means that σ(EL) ⊂ EL which is one characterization of a normal extension∎

It is clear that normal extensions are really *good and desirable* (kind of money, looks, and fame, but in the algebraic word 😄), so it will be important to know if we can extend a finite extension to make it normal. Let E/F be a finite extension, a field N containing E is said to be a normal closure over F if:

**N/F is a normal extension**.- If L is a proper subfield of N containing E, then L is not a normal extension of F. (Based on: Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.)

- Let E/F be a finite extension. Then, (i) there exists a normal closure N of E over F Futhermore (ii), if E’/F is a finite extension such that there is a F-isomorphism Φ: E → E’, and if N’ is a normal closure of E’ over F, then there is a F-isomorphism ψ: N → N’ such that the following diagram commutes.

Proof.

(i) Let E/F be a finite extension ⇒ Let {α_{1}, α_{2}, ··· α_{n}} be a basis for E over F. Each α_{i} is algebraic over F, then there exist a minimum (irreducible) polynomial with coefficients in F, let’s say m_{i}. Let N be the splitting field for m = m_{1}m_{2}···m_{n} over F (that is, by adjoining to F all the roots of all the m_{i}) ⇒ [A finite extension is normal iff it is the splitting field for some polynomial] Hence N/F is normal.

Let L be a subfield of N containing E, and suppose that L/F is normal ⇒ For each i, 1 ≤ i ≤ n, L (L ⊇ E) contains, at least, one root of m_{i}, namely α_{i} ⇒ [Equivalence of Definition of Normal Extensions. Every irreducible polynomial f ∈ F[x] with at least one root in L splits completely in L[x]] L contains all the roots of all m_{i} ⇒ [L is a subfield of N] L = N ∎

(ii) {α_{1}, α_{2}, ··· α_{n}} is a basis for E over F ⇒ every element of E (e ∈ E) has a unique expression as a_{1}α_{1} + a_{2}α_{2} + ··· + a_{n}α_{n} where a_{1}, a_{2}, ···, a_{n} ∈ F. Let e’ = Φ(e) be an arbitrary element of E’, e’ = Φ(e) = Φ(a_{1}α_{1} + a_{2}α_{2} + ··· + a_{n}α_{n}) = a_{1}Φ(α_{1}) + a_{2}Φ(α_{2}) + ··· + a_{n}Φ(α_{n}), therefore every element of E’ can be written uniquely as a linear combination of {Φ(α_{1}), Φ(α_{2}), ···, Φ(α_{n})}, that is, this is a basis for E’ over F.

Besides, the isomorphism Φ ensures that ∀i, 1 ≤ i ≤ n, the minimal polynomial of Φ(α_{i}) is Φ’(m_{i}) where Φ’ is the canonical extension of Φ to the polynomial rings E[x] → E’[x]. Since N’/E’ is normal by our theorem’s assumption, Φ(α_{i}) ∈ E’ ∀i ⇒ N’ must contain all the roots of all Φ’(m_{i}), and must indeed be a splitting field of Φ’(m_{1})Φ’(m_{2})···Φ’(m_{n}). The existence of the F-isomorphism ψ follows from the extension theorems, more specifically, 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 ∎

- Let K ⊆ L ⊆ F be finite extensions. Suppose K/F is a normal extension (⇒ K/L is normal). Then, L/F is a normal extension if and only if σ(L) ⊆ L ∀σ ∈ Gal(K/F).

Proof.

⇒) Suppose L/F is normal, let σ ∈ Gal(K/F).

Reclaim: One of the conditions of a normal extension is **σ: K → $\bar F$ is an F-homomorphism of fields, then σ(K) ⊆ K**.

Let’s restrict σ to L, σ|_{L}: L → K ⊆ $\bar L$, L/F is normal ⇒ σ(L) ⊆ L

⇐) Suppose σ(L) ⊆ L ∀σ ∈ Gal(K/F). Let τ be an arbitrary F-homomorphism, τ: L → $\bar L$, τ(L) ⊆ L ↭ L/F is normal?

Let τ be an arbitrary map τ: L → $\bar L$, we must check τ(L) ⊆ L. Notice that τ is an F-homomorphism, and [By extension theorem, L/K algebraic extension, **every field homomorphism Φ: L → C, where C is an algebraically closed field, can be extended to a homomorphism K → C**] we can extended it to a σ: K → $\bar L$.

Since K/F is normal ⇒ [ If σ: K → $\bar \mathbb{F}$ is an F-homomorphism of fields, then σ(K) ⊆ K, therefore σ induces an F-automorphism from K to itself σ: K → K. In fact, σ(K) = K -surjective-] σ(K) = K ⇒ σ: K → K, σ ∈ Gal(K/F) ⇒ [By assumption, **σ(L) ⊆ L ∀σ ∈ Gal(K/F)**] σ(L) ⊆ L ⇒ [σ is an extension of τ, σ|_{L} = τ] τ(L) ⊆ L ⇒ L/F is normal∎

- Let K ⊆ L ⊆ F be finite extensions. If K/F, K/L, L/F are all normal extensions. Then, we have an exact sequence of groups, 1 → Gal(K/L) → Gal(K/F) → Gal(L/F) → 1, that is, a sequence of group homomorphisms such that
**the image of one group homomorphism equals the kernel of the next**. More specifically, (1) ψ is injective, (2) Img(ψ) = Ker(φ) and (3) φ is surjective.

Proof.

(1) ψ is injective

Let σ ∈ Gal(K/L), σ: K → K, homomorphism that fixes L pointwise, σ(α) = α ∀α ∈ L ⇒ In particular, σ fixes F pointwise, σ(α) = α ∀α ∈ F, so σ ∈ Gal(K/F) ⇒ ψ(σ) = σ, and obviously ψ is injective because if σ is not the trivial map, that is, the identity, ψ(σ) will not be the identity.

(2) Img(ψ) = Ker(φ)

Let σ ∈ Gal(K/F), σ: K → K. Let’s restrict σ to L, σ|_{L}: L → K, but since L/F is normal [Let K ⊆ L ⊆ F be finite extensions. Suppose K/F is a normal extension ⇒ L/F is a normal extension if and only if σ(L) ⊆ L ∀σ ∈ Gal(K/F).] σ|_{L}(L) ⊆ L, and φ(σ) = σ|_{L}.

σ ∈ im(ψ) ⇒ σ ∈ Gal(K/L), and σ|_{L} = identity ⇒ φ(σ) = σ|_{L} = identity ⇒ [A map φ is a homomorphism between Galois groups, so **the “0” element is the identity map, let’s call it “id” with id(x) = x for all x in the base field**] σ ∈ Ker(φ) ⇒ im(ψ) ⊆ Ker(φ)

Conversely, let σ ∈Gal(K/F), σ ∈ Ker(φ) ⇒ φ(σ) = σ|_{L} = identity ⇒ σ ∈ Gal(K/L) ⇒ ψ(σ) = σ ⇒ σ ∈ im(ψ) ⇒ Ker(φ) ⊆ im(ψ) ⇒ Ker(φ) = im(ψ)

(3) φ is surjective.

Let σ ∈ Gal(L/F) ⇒ [By extension theorem] we can extend σ to K, say τ:K → $\bar L = \bar K$. Since K/F is normal ⇒ τ(K) = K ⇒ τ ∈ Gal(K/F) ⇒ φ(τ) = σ ⇒ φ is surjective.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn, Andrew Misseldine, and MathMajor, YouTube’s channels.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
- Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.