关于伽罗瓦理论中分裂域固定域引理证明及相关前提的疑问

Hey there! Let's work through your question about this key lemma in Galois Theory step by step.

First, let's restate the lemma clearly for context:

Given a field extension (F \subset E), where (E) is the splitting field over (F) of a separable polynomial (p(x) \in F[x]), then (E_G = F). Here:

• (G = G(E/F)) is the Galois group of the extension: (G(E/F) = { \sigma: E \to E \mid \sigma \text{ is an automorphism and } \sigma(a) = a \text{ for all } a \in F })
• (E_G) is the fixed field of (G) in (E): (E_G = { a \in E : \sigma(a) = a \text{ for every } \sigma \in G })

You're trying to prove this lemma by showing that if (\alpha \in E) and (\alpha \notin F) (and since (E) is the splitting field of (p(x)), (p(\alpha) = 0) automatically), then there exists some (\sigma \in G(E/F)) such that (\sigma(\alpha) = \beta), where (\beta) is a conjugate root of (\alpha). You mentioned this approach relies on a key assumption being true—let's unpack that assumption and why it holds.

The Key Precondition You Need #

The critical premise here is that for any two conjugate roots of an irreducible separable polynomial over (F), there exists an element of the Galois group that maps one root to the other. Here's why this is valid:

• Since (p(x)) is separable, it factors into distinct irreducible polynomials over (F): (p(x) = q_1(x)q_2(x)\dots q_k(x)), where each (q_i(x)) is irreducible and has no repeated roots.
• If (\alpha \notin F), it must be a root of some (q(x)) (one of the irreducible factors) with (\deg q(x) \geq 2). Since (q(x)) is separable, it has at least two distinct roots (\alpha) and (\beta).
• There's a natural field isomorphism (\sigma_0: F(\alpha) \to F(\beta)) that sends (\alpha) to (\beta) and fixes every element of (F) (this comes from the fact that (F(\alpha) \cong F[x]/(q(x)) \cong F(\beta)) for irreducible (q(x))).
• Because (E) is the splitting field of (p(x)) over (F), this isomorphism (\sigma_0) can be extended to an automorphism (\sigma: E \to E) that still fixes (F)—so (\sigma \in G(E/F)), and (\sigma(\alpha) = \beta \neq \alpha).

Putting It All Together for the Proof #

With that precondition confirmed, the lemma proof falls into place via contradiction:

• Suppose for contradiction that (E_G \neq F), so there exists some (\alpha \in E_G) with (\alpha \notin F).
• As above, (\alpha) is a root of an irreducible separable factor (q(x)) of (p(x)) with (\deg q(x) \geq 2), so there's another distinct root (\beta) of (q(x)).
• We can find (\sigma \in G(E/F)) such that (\sigma(\alpha) = \beta \neq \alpha), but this contradicts (\alpha \in E_G) (since elements of (E_G) are fixed by all elements of (G)).
• Therefore, our assumption is wrong, so (E_G \subset F). We already know (F \subset E_G) (every element of (F) is fixed by (G)), so (E_G = F).

备注：内容来源于stack exchange，提问作者SEHYUN YUK