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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].

Image 

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].

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.

n o r m a l F L | F N o t n o n r e m c a e l s s a r i l y n o r m a l , ( 2 ) ( 2 , i )

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.

n o r m a l | ( ( n 2 2 o , ) t i n ) o m r s i m p s a . s l f i i n e g l d n o o n f - r x e ^ a 4 l - r 2 o o t s

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.

n n o o t r m a l | | ( ( n n 2 o 2 o ) r ) r m m a a l l , , d d e e g g r r e e e e 2 2

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.

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.)
n o r m a l N E F \ p L r o p e r s u b f i e l d
i d F F - - Φ E E ' - - ψ N N '

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

n o r m a l K L F | | n n o o r r m m a a l l σ ( L ) L σ G a l ( 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?

K L σ / L τ , τ ( L ) L ?

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∎

K L F 1 G a l ( σ K / L ) ψ G a l ( σ K / F ) φ G σ a | l L ( L / F ) 1

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.

K | L F σ σ | K L L L σ G a l ( K / F )

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.

Bibliography

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