Galois extension II. Exercises.

Recall

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 = K^Gal(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.  

Dear Math, stop asking me to find your x. He’s not coming back, Anonymous.

Corollary. F ⊆ L ⊆ K are field extensions, K/F Galois ⇒ K/L is Galois.

Remark: It is not true that L/F is Galois, e.g., $\mathbb{Q}(\sqrt[3]{2},w)⊆\mathbb{Q}(\sqrt[3]{2})⊆ℚ$, $\mathbb{Q}(\sqrt[3]{2},w)/ℚ$ is Galois, but $\mathbb{Q}(\sqrt[3]{2})/ℚ$ is not Galois.

Proof: K/F Galois ⇒ K/F is both normal and separable ⇒ K/L is both normal and separable ⇒ K/L is Galois.

Lemma. Let K/F be a finite extension, then |Gal(K/F)| divides [K : F]

Proof.

F ⊆ K^Gal(K/F) ⊆ K are field extensions ⇒ [K : F] = [K : K^Gal(K/F)][K^Gal(K/F) : F] ⇒ [ [K : K^Gal(K/F)] = |Gal(K/F)|] ⇒ [K : F] = |Gal(K/F)|[K^Gal(K/F) : F] ⇒ |Gal(K/F)| divides [K : F]∎

Lemma. Every finite separable field extension K/F can be extended to a Galois extension.

Proof.

Let K/F be a finite separable field extension, K = F(α₁, α₂, ···, α_n), let f_i be the distinct irreducible polynomial of α_i over F, and [K/F is separable] each f_i is separable and if i≠j, f_i and f_j have no common roots in the algebraic closure (two distinct monic irreducible polynomials are coprime) ⇒ f=f₁···f_n is separable, too.

Let L be the splitting field of f over K, so we have F ⊆ K ⊆ L field extensions ⇒ [L is the splitting field of f separable over F] L is Galois over F∎

Fact: A Galois extension is defined as an extension which is both separable and normal. However, a normal extension does not need to be Galois, e.g., F = $\mathbb{F_p}(t)$, the field of fractions of R = $\mathbb{F_p}[t]$ which is a principal integral domain, p prime, t is a variable. Let’s consider f(x) ∈ F[x] given by f(x) = x^p -t, it is irreducible by Eisenstein’s criterion. Let K be the splitting field of f over F. Then, f has only one root in K.

Suppose α ∈ K be a root of f, α^p = t. Since Char(F) = Char(K) = p: (x -α)^p = [Because Char(F) = Char(K) = p, all the other intermediate factors are zero] x^p - α^p = x^p -t [A polynomial ring over a field is both a principal ideal domain (every ideal is principal) and a UFD, this factorization is unique] = f(x). f is irreducible, too.

f is irreducible over F: As this factorization f(x) = (x-α)^p is unique, its suffices to show that α is not in F (α^p = t). Let’s suppose α = $\frac{p(t)}{q(t)}: (\frac{p(t)}{q(t)})^p = t ⇒ p(t)^p=t·q(t)^p$ ⇒ [Let’s derivate, Char(F) = Char(K) = p] LHS = p·p(t)^p-1 = 0, RHS = q(t)^p + t·p·q(t)^p-1 = q(t)^p ⇒ [RHS = LFS] q(t)^p = 0 ⇒ q(t) = 0 ⇒ t = 0⊥

Therefore, f has only one root, say α, in K. K/F is normal because K is the splitting field of x^p -t, but K/F is not separable because α is not separable over F. The irreducible polynomial of α over F is f(x) = x^p -t, and f(x) does not have distinct roots in K ⇒ K/F is not Galois.

An alternative proof is that there exist only one F-automorphism of K, namely σ = identity (every automorphism sends α to α), σ: K → K, σ(α) = α ⇒ |Gal(K/F)| = 1 < [K : F] = p.

Exercises

• Let K be the field of rational functions in one variable over ℂ, i.e., K = ℂ(t) = {$\frac{f(t)}{g(t)}|f(t), g(t) ∈ ℂ[t], g(t) ≠ 0$}, σ₁: K → K, σ₁(t) = it, σ₂: K → K, σ₂(t) = t^-1 = ¹⁄_t.

σ₁: t → it → i·it=-t → -it → t, ord(σ₁) = 4, ord(σ₂) = 2, and σ₂σ₁ = σ₁³σ₂. Futhermore, G = ⟨σ₁, σ₂⟩ ≋ D₄, |D₄| = 8 ⇒ [K : K^G] = |G| = 8

Consider t⁴ + t^-4 ∈ K. σ₁(t⁴ + t^-4) = ((it)⁴ + (it)^-4) = t⁴ + t^-4, σ₂(t⁴ + t^-4) = t⁴ + t^-4, t⁴ + t^-4 is fixed by every element of G, ℂ(t⁴ + t^-4) ⊆ K^G. We want to prove equality.

What is the irreducible polynomial of t ∈ K over K^G (K/K^G is Galois)?

[By the previous theorem f = (x -α₁)(x -α₂)··· (x -α_r) ∈ K[x] is the irreducible polynomial of α over F] So to find the irreducible polynomial of t over K^G, let’s look at the orbit of t under G-action

orbit of t: t, (σ₁(t)) it, (σ₁²(t)) -t, -it, (σ₂(t)) t^-1, (σ₂(σ₁(t))) it^-1, -t^-1, -it^-1, 8 distinct elements in the orbit of t under G-action ⇒ the irreducible polynomial of t over K^G is (x -t)(x -it)(x +t)(x +it)(x -t¹)(x-it^-1)(x +t^-1)(x+it^-1) = [Please notice that (x -t)(x -it)(x +t)(x +it) = (x²-t²)(x²+t²) = (x⁴ -t⁴)] (x⁴ -t⁴)(x⁴ -t^-4) = x⁸ -(t⁴+t^-4)x⁴ +1 ∈ $\mathbb{C}(t^4+t^{-4})[x]$ ⇒ [An element α ∈ E is algebraic over F if there exists a non-constant polynomial p(x) ∈ F[x] such that p(α)=0] t is algebraic over K^G ⇒ K = ℂ(t) = K^G(t) = F(t)

[K : F] = [F(t) : F] = degree of irreducible polynomial of t over F ≤ 8. However, we already know that [K : F] ≥ [K : K^G] = |G| = 8 ⇒ [K : F] = 8 ⇒ K^G = F = $\mathbb{C}(t^4+t^{-4})$

Theorem. If a polynomial has degree 2 or 3 and has no roots over a field F, then f is irreducible in F[x]. In other words, a degree 2 or 3 polynomial over a field is reducible if and only if it has a root.

Counterexample. You do need a field, e.g., the polynomial 3(x²+1) ∈ ℤ[x]. It has no roots in ℤ but 3·(x² + 1) is a factorization. Another is 4x² -4x +1 = (2x -1)(2x -1) is reducible over ℤ, but no integer zeros.

A polynomial with coefficients in a unique factorization domain R is irreducible over R if it is an irreducible element of the polynomial ring, that is, it is not invertible, not zero, and cannot be factored into the product of two non-invertible polynomials with coefficients in R, e.g., 3 and x²+1 are non-invertible polynomials with coefficients in ℤ[x]

Proof. Every first degree cx + d has a root in F, namely -c^-1d. Therefore, if f = rs, then neither r nor s can have degree 1 ⇒ Since f has degree 2 or 3, either r or s must have degree 0 ⇒ r or s is constant ⇒ f is irreducible.

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.