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Author |
业余的业余
Post time 2023-12-3 02:44
Galois Theory
Tom Leinster, University of Edinburgh
Version of 31 March 2023
第99页定理7.1.5 大致等价$\newcommand{\SF}{\mathrm{SF}}$
Let $M:K$ be a field extension. Then \[
M=\SF_K(f)\text{ for some nonozero }f\in K[t]\iff M:K\text{ is finite and normal}.
\]
令 $M:K$ 为域扩张，那么\[
M\text{是} f\in K[t] \text{的分裂域} \iff M:K\text{ 有限且正规}.
\]
看看他的证明，我们只关心 $\Rightarrow$ 方向。
For $\Rightarrow$, take a nonzero $f\in K[t]$ such that $M=SF_K(f)$. Write $\alpha_1,\cdots,\alpha_n$ for the roots of $f$ in $M$. Then $M=K(\alpha_1,\cdots,\alpha_n)$. Each $\alpha_i$ is algebraic over $K$ (since $f\ne 0$), so by Proposition 5.2.4, $M:K$ is finite. （好吧，这个其实我们没那么关心,也比较显然).
We now show that $M:K$ is normal, which is the most substantial part of the proof. Let $\delta\in M$, with minimal polynomial $m\in K[t]$. Certainly $m$ splits in $SF_M(m)$, so to show that $m$ splits in $M$, it is enough to show that every root $\epsilon$ of $m$ in $SF_M(m)$ lies in $M$.
Since $m$ is a monic irreducible annihilating polynomial of $\epsilon$ over $K$, it is the minimal polynomial of $\epsilon$ over $K$. Hence by Theorem 4.3.16， there is an isomorphism $\theta:K(\delta)\to K(\epsilon)$ over $K$ such that $\theta(\delta)=\epsilon$. Now observe that:
- $M=\SF_{K(\delta)](f)$, by Lemma 5.2.14(ii);
- $K(\alpha_1,\cdots,\alpha_n,\epsilon)=\SF_{K(\epsilon)}(f)$, by Lemma 6.2.14(i);
- $\theta_*f=f$, since $f\in K[t]$ and $\theta$ is a homomorphism over $K$.
So we can apply Proposition 6.2.11, which implies that there is an isomorphism $\varphi:M\to K(\alpha_1,\cdots,\alpha_n,\epsilon)$ extending $\theta$. It is an isomorphism over $K$, since $\theta$ is.
Since $\delta\in K(\alpha_1,\cdots,\alpha_n)$ and $\varphi$ is a homomorphism over $K$, we have $\varphi(\delta)\in K(\varphi(\alpha_1),\cdots,\varphi(\alpha_n))$. Now $\varphi(\delta)=\theta(\delta)=\epsilon$, so $\epsilon\in K(\varphi(\alpha_1),\cdots,\varphi(\alpha_n))$. Moreover, for each $i$ we have $f(\varphi(\alpha_i))=0$ and so $\varphi(\alpha_i)\in \{\alpha_1,\cdots,\alpha_n\}$. Hence $\epsilon\on K(\alpha_1,\cdots,\alpha_n)=M$, as required.
牵涉甚广。不过大概可以看出思路与我贴出的第一个相类。这大概是比较近的伽罗华理论方面的作者喜欢采用的方式。 |
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Author: 业余的业余
abababa
Post time 2023-12-2 19:14
。