# Normal extensions 2

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:

1. 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.

2. Every irreducible polynomial f ∈ F[x] with at least one root in K, splits (into a product of linear factors) completely in K[x].
3. K is the splitting field of a polynomial f ∈ F[x].

# Facts

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 α1,···, αn∈ 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(α1,···, αn), where α1,···, αn are the roots of f. Futhermore, if f1(x),…, fn(x)∈ F[x], then there is a splitting field for {f1(x),…, fn(x)} over F, namely the splitting field of the product f = f1(x)···fn(x). Notice: We know there is a field E ⊇ F such that f splits over L, let α1, ···,αn be the roots of f. Then F(α1, ···,αn) 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 x4-2 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 - e1)(x - e2)···(x - ed) = (x - l1)(x - l2)···(x - ld). Since each of these factorizations of f are in F[x], which is a UFD, these li are just a reordering of the mj, 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 ai ∈ E, bi ∈ 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:

1. N/F is a normal extension.
2. 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 {α1, α2, ··· αn} be a basis for E over F. Each αi is algebraic over F, then there exist a minimum (irreducible) polynomial with coefficients in F, let’s say mi. Let N be the splitting field for m = m1m2···mn over F (that is, by adjoining to F all the roots of all the mi) ⇒ [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 mi, namely αi ⇒ [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 mi ⇒ [L is a subfield of N] L = N ∎

(ii) {α1, α2, ··· αn} is a basis for E over F ⇒ every element of E (e ∈ E) has a unique expression as a1α1 + a2α2 + ··· + anαn where a1, a2, ···, an ∈ F. Let e’ = Φ(e) be an arbitrary element of E’, e’ = Φ(e) = Φ(a1α1 + a2α2 + ··· + anαn) = a1Φ(α1) + a2Φ(α2) + ··· + anΦ(αn), therefore every element of E’ can be written uniquely as a linear combination of {Φ(α1), Φ(α2), ···, Φ(αn)}, 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(αi) is Φ’(mi) 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(αi) ∈ E’ ∀i ⇒ N’ must contain all the roots of all Φ’(mi), and must indeed be a splitting field of Φ’(m1)Φ’(m2)···Φ’(mn). 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.