Mathematics is like love; a simple idea, but it can get complicated, Anonymous.
Basically, the fundamental theorem of Galois theory says that you can tell a lot about a field extension by looking at its Galois group. More specifically, it says that for a finite and Galois field extension K/F there is a one-to-one correspondence between its intermediate fields and the subgroups of its Galois group. A finite extension K/F is Galois iff the order of the Galois group equals the degree of the extension, i.e., [K : F] = |Gal(K : F)|.
Notice that Galois’s fundamental theorem does not hold for all extensions. The easiest nontrivial extensions are the algebraic extensions, and more specifically, simple extensions (a field extension which is generated by the adjunction of a single element, called a primitive element), i.e., K = F(a). Let a ∈ K be algebraic over F, let f(x) be its irreducible polynomial, then F(a) = F[a] ≋ F[x]/⟨f(x)⟩, [F[a] : F] = deg f(x) = d, and a basis of F(a) as an F-vector space is {1, a, a^{2}, ···, a^{d-1}}, where d = deg(f), that is, a simple extension's degree equals the degree of the primitive element's minimal polynomial.
Fact. Suppose α is algebraic over F with minimal polynomial f(x) and K = F(α) ⇒ ∀ σ ∈ Aut(K/F)=Gal(K/F), σ(α) is also a root in f(x). Conversely, if β is any other root of f(x) in K ⇒ ∃! τ ∈ Aut(K/F)=Gal(K/F) with τ(α) = β. Therefore, |Gal(K/F)| is equal to the number of roots of f in K, and is (in particular) finite and at most [K : F].
If α_{1}, α_{2}, ···, α_{s} are the distinct roots of f in K. ∀σ ∈ Gal(K/F), σ(α_{i}) = α_{j} and we know that |Gal(K/F)| = s. Futhermore, [K : F] = deg(p), so [K : F] = |Gal(K/F)| if and only if deg(p) is equal to s, the number of distinct roots of the minimal polynomial of α over F.
Certainly, s ≤ deg(p), but they may not be equal because the irreducible polynomial might not split completely in K, e.g., F = ℚ, α = $\sqrt[3]{2}$, p(x) = x^{3} -2 which has only one root in K = ℚ($\sqrt[3]{2}$), so [ℚ($\sqrt[3]{2}$) : ℚ] = 3 = deg(x^{3} -2) > 1 = Gal(K/Q). An algebraic extension is called normal if the minimal polynomial of every element split in K. Secondly, some of the roots of p in K may be repeated. An algebraic extension is called separable if the minimal polynomial of every element has no repeated roots in its splitting field. If we take any finite extension K : F that is both normal and separable, then it is indeed true that |Gal(K : F)| = [K : F].
An algebraic field extension M/F is normal if for all α ∈ M, the minimal polynomial of α splits in M.
Lemma. Let M/F be an algebraic extension. Then M/F is normal if and only if every irreducible polynomial over F either has no roots in M or splits completely in M.
Proof.
⇒) Suppose M/F normal, let p be an irreducible polynomial over F. If f has no roots in M we are done. If f has a root in M, say α ∈ M ⇒ [the minimal polynomial of α is defined as the monic polynomial of least degree among all polynomials in F[x] having α as a root] the minimal polynomial of α is f/c where c ∈ F is the leading coefficient of f ⇒ [M/F is normal] f/c splits completely in M ⇒ f splits completely in M.
⇐) Suppose every irreducible polynomial over F either has no roots in M or splits completely in M. Let α ∈ M, the minimal polynomial of α has at least one root in M, namely α, so it splits completely in M.
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:
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.
It is essential that f is irreducible, e.g., K = $ℚ(\sqrt{2})$, F = ℚ. K/F has the conditions of the theorem, f = (x-2)(x^{2}+1) ∈ ℚ[x] has a root in K, namely 2, but it does not split completely in K[x] because f is not irreducible.
Proof.
1 ⇒ 2) Let f ∈ F[x] be an irreducible polynomial with at least one root in K, say α ∈ K be a root of f in K. Of course, f splits completely in $\bar \mathbb{F}[x]$.
Let β be an arbitrary root of f, F ⊆ F(α) (Notice F(α)⊆ K) ≋ [f is irreducible] F[x]/⟨f(x)⟩ ≋ F(β) ⊆ $\bar \mathbb{F}$
σ: F(α)→F(β) is defined by $σ(a_0 + a_1α + ··· + a_{n-1}α^{n-1})=σ(a_0)+σ(a_1)β+···+σ(a_{n-1})β^{n-1}$
Therefore, an extension σ: K → $\bar \mathbb{F}$ exists by the extension theorem ⇒ [1 Applies] σ(K) ⊆ K, α ∈ K, and β = σ(α) ∈ K. In other words, every root of f is in K ⇒ f splits completely over K.∎
2 ⇒ 3) Let K/F be a finite extension, therefore K can be written as K = F(α_{1},···, α_{n}) where α_{1},···, α_{n} ∈ K. Suppose that we need all α_{i} to generate K, i.e., we cannot remove any α_{i} to generate K.
A finite extension is algebraic ⇒ Consider f_{i} the irreducible polynomial of every α_{i} over F (Every f_{i} has a root in K and f_{i} is irreducible), α_{i} ∈ K ⇒ [By 2] f_{i} splits completely over K[x].
Consider f = f_{1}f_{2}···f_{n}. Then, f splits completely over K[x] because every singular f_{i} splits completely over K[x] and K = F(α_{1},···, α_{n}), i.e., the smallest subfield of K containing all the roots of f ∴ K is the splitting field of f over K∎
Recall that a splitting field of a polynomial over a field K is the smallest field extension of that field over which the polynomial splits, i.e., decomposes into linear factors.
3 ⇒ 1) Let K be the splitting field of a polynomial f ∈ F[x] over F. f(x) = p_{1}(x)·p_{2}(x) ··· p_{n}(x), where p_{i}(x) has all their roots in K, so suppose that the roots of all p_{i} are α_{1}, ···, α_{n}, K = F(α_{1}, ···, α_{n}).
Let σ: K → $\bar \mathbb{F}$ be an F-homomorphism. We already have demonstrated that ∀α_{i}: σ(α_{i}) ∈ {α_{1}, α_{2}, ···, α_{n}}
Therefore, ∀i σ(α_{i}) ∈ K ⇒ [K = F(α_{1}, ···, α_{n})] σ: K → $\bar \mathbb{F}$, σ(K) ⊆ K because every polynomial in K is a polynomial in α_{i} con coefficients in F.∎
Let E be an algebraic extension of a field F contained in an algebraic closure $ \bar \mathbb{F}$ of F. The following statements are equivalent:
An extension E/F satisfying one (and hence all) of these conditions is called normal.
Proof.
i ⇒ ii)
Let α ∈ E ⇒ [E/F algebraic] α is algebraic over F. Let p_{α}(x) be its minimal polynomial over F. ⇒ [By assumption, α ∈ E is a root of p_{α}(x) (i)] p_{α}(x) splits into linear factor in E ⇒ E is the splitting field of a family of polynomials {p_{α}(x)}_{α ∈ E}
ii ⇒ iii)
Let E be the splitting field of a family of polynomials {f_{i}(x)}_{i ∈ γ}, we may write f_{i}(x) = u$\prod_{i=1}^n (x-u_i)$ for some u, u_{1}, ···, u_{n} ∈ E. If σ: E → $\bar \mathbb{F}$ is an embedding (1-1, homomorphism) that keeps each element of F fixed ⇒ Any root of f_{i} must be sent to another root by σ (homomorphism), σ(u_{i}) ∈{u_{1},···, u_{n}} but since σ is injective, it must simple permute the roots of f. As E is generated over F by the roots of these polynomials, we obtain σ(E) ⊆ E ⇒ σ is onto.
iii ⇒ i) Let p(x) be an irreducible polynomial over F that has a root α ∈ E. Suppose β ∈ $\bar \mathbb{F}$ be another root of p(x).
Claim. β ∈ E, i.e, all roots of p(x) are in E, thus the polynomial splits into linear factors in E.
As α, β are both roots of p(x) ⇒ We have an F-isomorphism F(α) ≋ F[x]/⟨p(x)⟩ ≋ F(β). Let σ: F(α) → F(β) be the isomorphism, σ|_{F} = id|_{F}, σ(α) = β. E is algebraic over F(α), and by the extension theorem, σ can be extended to an embedding σ’: E → $\bar \mathbb{F}$ ⇒ [By our assumption (iii)] σ’ is an automorphism, and maps E onto E ⇒ σ’(α) = [σ’ is an extension of σ, σ’|_{F(α)} = σ] σ(α) = β ∈E
ℂ is a normal extension of ℝ because every polynomial in ℝ splits completely into linear factors in ℂ.
ℝ is not a normal extension over ℚ because x^{3} -2 ∈ ℚ[x], one of its root 2^{1/3} ∈ ℝ, but the other two roots, as they are both complex, does not belong to ℝ.
$ℚ(\sqrt{2})$ is a normal extension of ℚ since $ℚ(\sqrt{2})$ is a splitting field of p = x^{2} -2.
K = $ℚ(\sqrt[3]{2})$, F = ℚ. It is not normal. [If f ∈ F[x] is an irreducible polynomial with at least one root in K, then f splits (that is, it splits as a product of linear factors) completely in K[x]] because x^{3} -2 ∈ ℚ[x] is irreducible, it has a root in K, namely $\sqrt[3]{2}$, but does not split completely (it does not have the non-real cubic roots of 2). There are more examples here.
Let α ∈ K \ ℚ ⇒ [K is an extension of ℚ of degree 2] Claim: K = ℚ(α). [ℚ(α) : ℚ] ≥ 2 because α ∈ K \ ℚ.
ℚ ⊆ ℚ(α) ⊆ K, then [K : ℚ] = [By hypothesis] 2 = [K : ℚ(α)][ℚ(α) : ℚ] = 1·2 and [K : ℚ(α)] = 1 (otherwise, there is a contradiction ⊥ [ℚ(α) : ℚ] = 1, but α ∈ K \ ℚ).
Let f = x^{2} + bx + c ∈ ℚ[x] be the irreducible polynomial of α (it is obviously of degree 2). Suppose that α, β are roots of f in ℂ (By the Fundamental Theorem of Algebra, ℂ is algebraically closed). Then α + β = -b ⇒ β = -b -α ∈ K (b ∈ ℚ ⊆ K, α ∈ K). Therefore, K is the splitting field of f over ℚ, i.e., K is normal over F = ℚ (Observe that α ≠ β because F = ℚ has characteristic zero).
Let f(x) be an irreducible polynomial over a field F. If the field (e.g., ℚ) has characteristic zero, then f(x) has no multiple zeros.
Let σ: K → K, σ(α) = β ≠ α, σ is a ℚ-automorphism of K, and σ ≠ identity ⇒ Gal(K/Q) = {1, σ} ⇒ |Gal(K/Q)| = 2 = [K : ℚ] ⇒ K/ℚ is Galois.