# 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 = KGal(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 ⊆ KGal(K/F) ⊆ K are field extensions ⇒ [K : F] = [K : KGal(K/F)][KGal(K/F) : F] ⇒ [ [K : KGal(K/F)] = |Gal(K/F)|] ⇒ [K : F] = |Gal(K/F)|[KGal(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, ···, αn), let fi be the distinct irreducible polynomial of αi over F, and [K/F is separable] each fi is separable and if i≠j, fi and fj have no common roots in the algebraic closure (two distinct monic irreducible polynomials are coprime) ⇒ f=f1···fn 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) = xp -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, αp = t. Since Char(F) = Char(K) = p: (x -α)p = [Because Char(F) = Char(K) = p, all the other intermediate factors are zero] xp - αp = xp -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 (αp = 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 xp -t, but K/F is not separable because α is not separable over F. The irreducible polynomial of α over F is f(x) = xp -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 = 1t.

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 : KG] = |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) ⊆ KG. We want to prove equality.

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

[By the previous theorem f = (x -α1)(x -α2)··· (x -αr) ∈ K[x] is the irreducible polynomial of α over F] So to find the irreducible polynomial of t over KG, 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, (σ21(t))) it-1, -t-1, -it-1, 8 distinct elements in the orbit of t under G-action ⇒ the irreducible polynomial of t over KG 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 KG ⇒ K = ℂ(t) = KG(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 : KG] = |G| = 8 ⇒ [K : F] = 8 ⇒ KG = 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.