Advanced Algebra I
splitting fields and normal extensions
In this section, we are going to prove the existence and uniqueness of splitting fields. And we introduce the notion of normal extension.
Proposition 0.1. Let K be a field. And S be a set of polynomial in K[x]. Then
(1) There is a splitting of S over K.
(2) Any two splitting field are isomorphic.
(3) If F1, F2 are two splitting fields in a fixed algebraic closure K, then F1 = F2.
Proof. In an algebraic closure K, we pick all roots of polynomials of S, say {ui}i∈I. Then K(ui)i∈I is a splitting field.
Let F1 and F2 be two splitting fields, one has an embedding σ : K → F2 = K. This embedding can be extended to ˜σ : F1 → F2 by the extension theorem. One can prove that image of ˜σ is in F2. Hence one has an injective homomorphism ˜σ : F1 → F2. Similarly there is another one ˜τ : F2 → F1. It’s easy to show that these give the isomorphism. ¤ Proposition 0.2. Let N be an algebraic extension over K contained in K. Then the following are equivalent:
(1) Any K-embedding σ : N → K induces an K-automorphism of N.
(2) N is a splitting field of some S ⊂ K[x] over K.
(3) Every irreducible polynomial in K[x] having a root in N splits in N.
Proof. For (1) ⇒ (2), (3), we prove that for every u ∈ N, with minimal polynomial p(x), then v ∈ N for every root of p(x). To this end, start with an isomorphism σ : K(u) → K(v). By extension theorem, one can extend it to an embedding N → K(v) = K. The embedding is an automorphism by (1). Thus, v = σ(u) ∈ N.
(3) ⇒ (2) is trivial.
For (2) ⇒ (1). Suppose that N is a splitting field of S over K. Let u be a root of f (x) ∈ S. Let σ : N → K be any K-embedding. It’s clear that σ(u) is a root of f (x), hence σ(u) ∈ N. Thus σ(N) ⊂ N. Since σ is injective and N/K is algebraic, σ is in fact an isomorphism. ¤ The property of being normal is not as well-behaved as begin alge- braic or finite. For example, it’s not preserve after ”extension”
Example 0.3. If F/E and E/K are normal, then F/K is not nec- essarily normal. For example, take F = Q(√4
2), E = Q(√
2), K = Q.
It’s easy to see that a degree 2 extension is always normal, however, Q(√4
2) is not normal over Q.
1
2
Also let’s consider K ⊂ E ⊂ F . Then F is normal over K implies that F is normal over E. But it doesn’t imply that E is normal over K. For example, take F = Q(√4
2, i), E = Q(√4
2), K = Q Being normal is preserved by ”lifting” and ”compositum”
Proposition 0.4. Let E, F be extensions over K and contained in a field L. If E/K is normal then EF/F is normal. Moreover, if both E/K, F/K are normal, then EF/K is normal.
Proof. In order to show that EF is normal over F , we look at F - embedding σ : EF → F . Since σ is identity on F , hence on K. By the extension theorem and the proof of the previous Proposition, one can show that σ|E is an automorphism. Hence σ(E) = E. It follows that
σ(EF ) = σ(E)F = EF.
Thus EF is normal over F .
Suppose now that E/K, F/K are normal. Let σ : EF → K be a K-embedding. We have that σ|E, σ|F are K-embeddings. One sees that σ(E) = E and σ(F ) = F by the normal assumption. If follows that
σ(EF ) = σ(E)σ(F ) = EF.
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