Theorem. Let K/F be a finite extension. The following statements are equivalent:

- K/F is Galois.
- F is the fixed field of Aut(K/F), i.e., F = K
^{Gal(K/F)} - 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]
- K/F is a normal, finite, and separable extension.
- K is the splitting field of a separable polynomial f ∈ F[x] over F.
- 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(α_{1}, α_{2}, ···, 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=f_{1}···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.

- Let K be the field of rational functions in one variable over ℂ, i.e., K = ℂ(t) = {$\frac{f(t)}{g(t)}|f(t), g(t) ∈ ℂ[t], g(t) ≠ 0$}, σ
_{1}: K → K, σ_{1}(t) = it, σ_{2}: K → K, σ_{2}(t) = t^{-1}=^{1}⁄_{t}.

σ_{1}: t → it → i·it=-t → -it → t, ord(σ_{1}) = 4, ord(σ_{2}) = 2, and σ_{2}σ_{1} = σ_{1}^{3}σ_{2}. Futhermore, G = ⟨σ_{1}, σ_{2}⟩ ≋ D_{4}, |D_{4}| = 8 ⇒ [K : K^{G}] = |G| = 8

Consider t^{4} + t^{-4} ∈ K. σ_{1}(t^{4} + t^{-4}) = ((it)^{4} + (it)^{-4}) = t^{4} + t^{-4}, σ_{2}(t^{4} + t^{-4}) = t^{4} + t^{-4}, t^{4} + t^{-4} is fixed by every element of G, ℂ(t^{4} + 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 -α_{1})(x -α_{2})··· (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, (σ_{1}(t)) it, (σ_{1}^{2}(t)) -t, -it, (σ_{2}(t)) t^{-1}, (σ_{2}(σ_{1}(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 -t^{1})(x-it^{-1})(x +t^{-1})(x+it^{-1}) = [Please notice that (x -t)(x -it)(x +t)(x +it) = (x^{2}-t^{2})(x^{2}+t^{2}) = (x^{4} -t^{4})] (x^{4} -t^{4})(x^{4} -t^{-4}) = x^{8} -(t^{4}+t^{-4})x^{4} +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(x^{2}+1) ∈ ℤ[x]. It has no roots in ℤ but 3·(x^{2} + 1) is a factorization. Another is 4x^{2} -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 x^{2}+1 are non-invertible polynomials with coefficients in ℤ[x]

Proof. Every first degree cx + d has a root in F, namely -c^{-1}d. 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.

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.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Fields and Galois Theory. Howie, John. M. Springer Undergraduate Mathematics Series.
- Fields and Galois Theory. Morandi. P., Springer.
- Fields and Galois Theory. By Evan Dummit, 2020.