JustToThePoint English Website Version
JustToThePoint en español

Maximize your online presence with our exclusive offer: Get a stunning hero banner, the hero you need and deserve, at an unbeatable price! Bew, 689282782, bupparchard@gmail.com

Automorphisms. Galois extensions and groups

We shall defend our island, whatever the cost may be. We shall fight on the beaches, we shall fight on the landing-grounds, we shall fight in the fields and in the streets, we shall fight in the hills. We shall never surrender!, Winston Churchill.

Recall

A field L is said to be an extension field, denoted L/F, of a field F if F is a subfield of L, e.g., ℝ/ℚ and ℂ/ℝ. The degree of L/F, denoted [L:F] = dimkL is the dimension of E considered as a vector space over, e.g., [ℂ : ℝ] = 2, [ℝ : ℚ] = ∞.

L/K, α is algebraic over K if α is the root of a polynomial p(x), with coefficients in the field K[x]. Otherwise, α is called transcendental. If α is algebraic, there is a smallest irreducible polynomial that it's a root of, and degree of α is the degree of such polynomial.

If X is any structured set, e.g., group, ring, field, topology, partially ordered set, etc. then an automorphism of X is a bijective function Φ: X → X such that preserves the structure of X, e.g., preserve group multiplication, addition, adjacency, etc.

Theorem. The set of all automorphisms of a field F, denoted or written as Aut(F), is a group under composition or mapping of functions. We refer to Aut(F) as the group of automorphisms of F.

Image 

Proof.

  1. Associativity. Composition of functions is always associative, ∀α, β, τ ∈ Aut(F), ((α∘β)∘γ)(x) = (α∘β)(τ(x)) = α(β(τ(x))) = α(β∘τ(x)) = (α∘(β∘τ))(x).
  2. The identity i ∈ Aut(F), i(x) = x ∀x∈ F is obviously an automorphism such that i ∘ α = α ∘ i ) = α, ∀α ∈ Aut(F).
  3. ∀α ∈ Aut(F), ∃α-1 defined as follows, α-1(x) is the unique z ∈ F: α(z) = x.

Let’s prove that α-1 ∈ Aut(F), ∀x, y ∈ F ∃z, t: α-1(x) = z, α-1(y) = t ⇒ [By definition] α(z) = x, α(t) = y ⇒ [α ∈ Aut(F)] α(z + t) = x + y ⇒ α-1(x) + α-1(y) = z + t = α-1(α(z + t)) = α-1(x + y). Mutatis mutandis, (α-1(x))(α-1(y)) = (α-1(xy)), thus α-1 ∈ Aut(F)∎

Let E be a field extension of F, then the subset of all automorphisms of E (σ: E → E) that fix F (σ(α) = α ∀α ∈ F), also called F-automorphisms of E, is a subgroup of Aut(F), denoted as Aut(E/F).

Proof:

Recall. The set of all automorphisms of a field F, denoted or written as Aut(F), is a group under composition or mapping of functions.

Since the identity fixes every element of E, in particular it fixes every element of F, id ∈ Aut(E/F).

Let σ, τ ∈ Aut(E/F), σ∘τ is an automorphism of E. It suffices to show that it fixes F ⇒ ∀α ∈ F, σ∘τ(α) = σ(τ(α)) = [τ∈Aut(E/F)] σ(α) =[σ∈Aut(E/F)] α ⇒

σ-1 is an automorphism of E. It suffices to show that it fixes F. ∀α ∈ F, α = σ-1(σ(α)) = [σ∈ Aut(E/F)] σ-1(α) ⇒ σ-1 ∈ Aut(E/F).∎

Recall Let K/F be a finite field extension. An element α ∈ K is separable over F if its irreducible polynomial over F is separable. An arbitrary polynomial over a field F is separable if its roots are distinct in an algebraic closure of F. A finite extension K/F is separable if every element of K is separable over F.

We say that E/F is a Galois extension if it is separable and normal, i.e., E is a splitting field for a polynomial or family of polynomials with no repeated roots.

If f(x) ∈ F[x] is an irreducible polynomial over a field F of characteristic zero (char(F)=0) or a finite field (|F| < ∞), then f(x) has no repeated roots and their extensions will always be separable.

Definition. Let E/F be a field extension. An automorphism of E is a ring isomorphism from E onto itself. The Galois group of E over F, written as Gal(E/F), Gal(E, F) or Gal(E : F), is the set of all field automorphisms of E which fix the base field F (i.e. take every element of F to itself). Gal(E/F) = {σ ∈ Aut(E): σ(α) = α ∀α ∈ F}.

If f(x) is a polynomial in F[x] and E is the splitting field of f(x) over F, we define the Galois group of f(x) to be G(E/F).

Example. Compute the group Gal(ℂ, ℝ).

σ ∈ Gal(ℂ, ℝ) ⇒ [By definition] σ(α) = α ∀α ∈ ℝ. Let σ(i) = j ⇒ j2 = (σ(i))2 = [σ ∈ Gal(ℂ, ℝ)] σ(i2) = σ(-1) = [-1 ∈ ℝ] -1 ⇒ [j2 = -1] j = ± i. Therefore, there are two options:

  1. j = i ⇒ ∀α ∈ ℂ, α = x +yi, σ(α) = σ(x + yi) = [σ ∈ Gal(ℂ, ℝ), σ automorphism and fixes ℝ] σ(x) + σ(y)σ(i) = x + yi ⇒ σ = id, the identity automorphism.
  2. j = -i ⇒ ∀α ∈ ℂ, α = x +yi, σ(α) = σ(x + yi) = [σ ∈ Gal(ℂ, ℝ), σ automorphism and fixes ℝ] σ(x) + σ(y)σ(i) = x - yi.

To prove that it is an automorphism, ∀α, β ∈ ℂ, α = x + yi, β = u + vi, σ(α + β) = σ((x +yi) + (u + vi)) = σ((x +u) + (y + v)i) = [By definition] (x + u) -(y +v)i = (x -yi) + (u -vi) = σ(α) + σ(β).

∀α, β ∈ ℂ, α = x + yi, β = u + vi, σ(α·β) = σ((x +yi)·(u + vi)) = σ((xu -yv) + (xv + yu)i) = [By definition] (xu - yv) -(xv +yu)i = (x -yi)·(u -vi) = σ(α)·σ(β).

Therefore, Gal(ℂ, ℝ) is the group {idC, σ} of order 2, where σ is the complex conjugation function: x + yi → x -yi. Gal(ℂ, ℝ) ≋ ℤ2. Futhermore, [ℂ : ℝ] = [i ∉ ℝ ( ⇒ [ℂ : ℝ] > 1), x2 + 1 is the irreducible polynomial of i of degree 2] 2 = |Gal(ℂ, ℝ)|. We will demonstrate that Theorem. 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]

Besides, α(i) = ±i, that is, σ permutes roots of f(x) = x2+1 (the irreducible polynomial of i).

Theorem. Let E be an extension of a field F and f(x) be a minimal F-polynomial of α ∈ E. Then, any automorphism σ of E leaving F fixed (σ ∈ Gal(E/F)) induces a permutation of the set of all zeros or roots of f(x) that are in E.

Proof.

Let f(x) = a0 + a1x + a2x2 + ··· anxn ∈ F[x]. Suppose α ∈ E is a root of f(x), f(α) = 0.

σ ∈ Gal(E/F), 0 = [σ ∈ Gal(E/F), σ is an automorphism] σ(0) = [α is a root of f(x)] σ(f(α)) = σ(a0 + a1α + a2α2 + ··· anαn) = [σ ∈ Gal(E/F), σ is automorphism that fixes F] a0 + a1σ(α) + a2[σ(α)]2 + ··· an[σ(α)]n ⇒ σ(α) is another root of f. Therefore, σ maps roots of f to roots of f, and this map must be a permutation of the roots of f(x) ∎

It is difficult for the reader to be able to grasp the far reaching consequences of this result.

Definition. Let E be an algebraic extension of a field F. We say that two elements in E (α, β ∈ E) are conjugate over F if they share the same minimal polynomial over F[x], e.g., in the field ℚ($\sqrt{2}$), $\sqrt{2}, -\sqrt{2}$ are conjugate over ℚ since they are both roots of the irreducible polynomial x2 -2.

If α, β ∈ E are conjugate over F, then the map σ: F(α) → F(β) which fixes F (σ = IdF) and send α to β, i.e., α → β can be shown to be a field isomorphism. σ can be extended to the algebraic closure of F, $σ: \bar F → \bar F$ and we will get a field automorphism that fixes the base field F, and it will permute the conjugates of the irreducible polynomial. Thus, a field automorphism induces a permutation on the conjugates of irreducible polynomials.

Example:

Let’s describe Gal($ℚ(\sqrt{2},i\sqrt{3}), ℚ$). $ℚ(\sqrt{2},i\sqrt{3})$ = {$a + b\sqrt{2}+ci\sqrt{3}+di\sqrt{6}$}. By the previous theorem, if σ ∈ Gal($ℚ(\sqrt{2},i\sqrt{3}), ℚ$) ⇒ [x2 -2 = 0, x2 + 3 = 0 are the irreducible polynomials of $\sqrt{2},~ and~ i\sqrt{3}$ respectively] $σ(\sqrt{2}) = ±\sqrt{2}, σ(i\sqrt{3})=±i\sqrt{3}$.

And therefore, there are only four possibilities: id, $σ(a + b\sqrt{2}+ci\sqrt{3}+di\sqrt{6}) = a - b\sqrt{2}+ci\sqrt{3}-di\sqrt{6}, τ(a + b\sqrt{2}+ci\sqrt{3}+di\sqrt{6}) = a - b\sqrt{2}-ci\sqrt{3}-di\sqrt{6}, θ(a + b\sqrt{2}+ci\sqrt{3}+di\sqrt{6}) = a - b\sqrt{2}-ci\sqrt{3}+di\sqrt{6}$.

Gal($ℚ(\sqrt{2},i\sqrt{3}), ℚ$) = {id, σ, τ, θ} ≋ K4, the Klein 4-group, a group of four elements, each of which is self-inverse, i.e., H1 = {id, σ}, H2 = {id, τ}, and H3 = {id, θ} ≤ Gal($ℚ(\sqrt{2},i\sqrt{3}), ℚ$).

Theorem. Let f(x) be a polynomial in F[x] (f(x) ∈ F[x]) and let E be the splitting field for f(x) over F. If f(x) has no repeated roots, the order of the Galois group equals the degree of the extension, |Gal(E/F)| = [E:F].

Proof. Credits: Maths StackExchange, Proof of Order of Galois Group equals Degree of Extension, @Noble Mushtak.

Let f(x) be a polynomial in F[x] with no repeated roots. Suppose E is the splitting field for f(x) over F and [E : F] = n. We are going to do induction on the degree of the extension of E/F, n.

If [E : F] = 1 ⇒ Clearly, E = F and the only possible automorphism is the identity ⇒ Gal(E/F) = {id} ⇒ |Gal(E/F)| = [E : F] = 1.

Now, let’s do strong induction on n. Assume that for all fields F, for any splitting field E of some polynomial f(x) ∈ F[x] such that [E : F] < n, |Gal(E/F)| = [E : F].

Let f(x) = p(x)q(x), where p(x) is an irreducible factor of f(x) over F[x] whose degree is r. We may assume r > 1; otherwise, f(x) splits over F and [E:F] = 1⊥

Let E be an extension field of F, f(x) ∈ F[x] has degree n ≥ 1. We say that f(x) splits in E (and E a splitting field for f(x)) if there are a, a1, ···, an such that f(x) = a(a - a1) ··· (x - an).

Since f(x) splits over E ⇒ p(x) splits over E and all the r roots of p(x) are in E, namely α1, α2, ···, αr

[Theorem (*). Let Φ: E → F be an isomorphism of fields. Let K be an extension 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 the root of the polynomial in F[x] obtained from p(x) under the image of Φ. Then, Φ extends to a unique isomorphism $\bar Φ: E(α) → F(β): \bar Φ(α)=β,~ and~ \bar Φ$ agrees with Φ on E]

∀1 ≤ i ≤ r, there is a unique isomorphism Φi: F(α1) → F(αi) such that Φi fixes F and Φi1) = αi

[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) be the corresponding polynomial under the isomorphism Φ. If K is a splitting field of p(x) and L is a splitting field of q(x), then Φ extends to an isomorphism ψ: K → L such that ψ agrees with Φ on E.]

Then, there exists an isomorphism ψi: E → E such that ψi agrees with Φi on F(α1).

Image 

Next, take some arbitrary automorphism σ ∈ Gal(E/F) ⇒ [Let E be a field extension of F and f(x) be a polynomial in F[x]. Then, any automorphism in Gal(E/F) defines a permutation of the roots of f(x) which lie in E.] so σ(α1) = αi for some 1 ≤ i ≤ r, σ(F) = F(by definition σ ∈Gal(E/F)) ⇒ σ(F(α1)) contains both F and αi ⇒ [F(α) is the smallest subfield of E containing both F and α] σ(F(α1)) ⊆ F(αi)

Now, consider σ-1(F(αi)), clearly σ-1(F) = F since σ fixes F. Besides, σ(α1) = αi ⇒ σ-1i) = α1 ⇒ σ-1(F(αi)) contains both F and α1 ⇒ σ-1(F(αi)) ⊆ F(α1) ⇒ F(αi) ⊆ σ(F(α1)) ⇒ [We have previously demonstrated that σ(F(α1)) ⊆ F(αi)] F(αi) = σ(F(α1))

σ restricted to F(α1) is an isomorphism from F(α1) to F(αi) such that σ fixes F and σ(α1) = αi. However, by the previous theorem (*) there is only one isomorphism which satisfies these conditions, and Φi, as defined below, is also an isomorphism such that Φi fixes F and Φi1) = αi ⇒ σ must agree with Φi on F(α1)

Let consider ψi-1σ. As it was previously stated, both σ and ψi agree with Φi on F(α1) ⇒ ∀x ∈ F(α1), $ψ_i^{-1}σ(x)=ψ_i^{-1}(Φ_i(x))$= [ψi and Φi agree on F(α1) and Φi(F(α1)) = F(αi) ⇒ ψi-1 and Φi-1 agree on F(αi)] $Φ_i^{-1}(Φ_i(x))=x$ ⇒ ψi-1σ is an automorphism of E which fixes F(α1), that is, ψi-1σ ∈ Gal(E/F(α1))

⇒ [If E is a finite extension of F and K is a finite extension of E, then K is a finite extension of F and [K : F] = [K : E][E : F].]

Thus [E : F] = [E : F(α1)][F(α1): F] ⇒ [E : F(α1)] = [E : F][F(α1): F] = n/r, thus [E : F(α1)] < n, and the induction hypothesis applies, |Gal(E/F(α1))| = |E : F(α1)| = n/r, we can label these elements as θ1, θ2,···, θnr.

As we have discussed earlier in this proof, ψi-1σ ∈ Gal(E/F(α1)) ⇒ ψi-1σ = θj for some 1 ≤ j ≤ nr. In other words, for any σ ∈ G(E/F), σ = ψiθj for some i, j such that 1 ≤ i ≤ r, 1 ≤ j ≤ nr ⇒ So there are r possibilities for ψi and n/r for θj, and therefore there are r·(n/r) = n possibilities for σ ⇒ |G(E/F)| = n = [E:F] ∎

Bitcoin donation

JustToThePoint Copyright © 2011 - 2024 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.

This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our use of cookies, in accordance with our Cookies Policy and Website Terms and Conditions of use.