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本帖最后由 hbghlyj 于 2023-8-13 23:57 编辑 sheet 6 Q6(iii)
6. Say \(E ⊆ F\), and \(L\) and \(M\) are intermediate fields (i.e. \(E ⊆ L, M ⊆ F\)). Let \(N:=L M\) denote the smallest subfield of \(F\) containing \(L\) and \(M\).
i) If \(L=E\left(α_{1}, …, α_{n}\right)\) then prove \(N=M\left(α_{1}, …, α_{n}\right)\).
ii) Now assume \(L / E\) and \(M / E\) are finite and normal. Prove \(N / E\) is finite and normal. (hint: splitting field). Next assume \(L / E\) and \(M / E\) are finite and Galois. Prove that \(N / E\) is finite and Galois.
iii) Prove that restriction of functions gives a natural injective group homomorphism from \(\operatorname{Gal}(N / E)\) to \(\operatorname{Gal}(L / E) × \operatorname{Gal}(M / E)\). Is it always surjective?
参考答案：solns
If \(g ∈ \operatorname{Gal}(N / E)\) then \(g(L)=L\) by 6.7 and hence the restriction of \(g\) to \(L\) is in \(\operatorname{Gal}(L / E)\). Similar for \(M / E\). So we get a map \(\operatorname{Gal}(N / E) → \operatorname{Gal}(L / E) × \operatorname{Gal}(M / E)\). This is easily checked to be a group homomorphism. It's injective because anything in the kernel fixes \(L\) and \(M\) pointwise, so fixes \(L M\) pointwise; but \(L M=N\).
It's not always surjective though – for example if \(L=M\) then it hardly ever is. More generally if \(L ∩ M ≠ E\) then there will be problems. However if \(L ∩ M=E\) then my guess is that the map is a bijection; however it's nearly midnight and so I think I'll leave this as an exercise for the interested reader!
又见 Keith Conrad The Galois Correspondence At Work Theorem 2.1
\[
\xymatrix{
&L_1L_2\ar@{-}[dl]\ar@{-}[dr]\\
L_1\ar@{-}[dr]&&L_2\ar@{-}[dl]\\
&K
}
\]
Theorem 2.1. Let $L_1$ and $L_2$ be Galois over $K$.
a) The embedding
$$
\operatorname{Gal}\left(L_1 L_2 / K\right) \hookrightarrow \operatorname{Gal}\left(L_1 / K\right) \times \operatorname{Gal}\left(L_2 / K\right)
$$
given by $\sigma \mapsto\left(\left.\sigma\right|_{L_1},\left.\sigma\right|_{L_2}\right)$ is an isomorphism if and only if $L_1 \cap L_2=K$. In particular, $\left[L_1 L_2: K\right]=\left[L_1: K\right]\left[L_2: K\right]$ if and only if $L_1 \cap L_2=K$.
b) The image of the embedding in part a is the set of compatible pairs of automorphisms: $\left\{\left(\tau_1, \tau_2\right) \in \operatorname{Gal}\left(L_1 / K\right) \times \operatorname{Gal}\left(L_2 / K\right): \tau_1=\tau_2\right.$ on $\left.L_1 \cap L_2\right\}$.
Proof. a) The embedding is an isomorphism if and only if $\left[L_1 L_2: K\right]=\left[L_1: K\right]\left[L_2: K\right]$, or equivalently $\left[L_1 L_2: L_2\right]=\left[L_1: K\right]$. We will show this equality occurs if and only if $L_1 \cap L_2=K$.
又见MSE帖子Converse of Galois translation theorem |
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