Lemma 9.16.6 (0EXL)—The Stacks project

Lemma 9.16.6. Let $L/K$ be a finite extension. Let $M/L$ be the normal closure of $L$ over $K$. Then there is a surjective map

\[ L \otimes _ K L \otimes _ K \ldots \otimes _ K L \longrightarrow M \]

of $K$-algebras where the number of tensors can be taken $[L : K]_ s \leq [L : K]$.

Proof. Choose an algebraic closure $\overline{K}$ of $K$. Set $n = [L : K]_ s = |\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K})|$ with equality by Lemma 9.14.8. Say $\mathop{\mathrm{Mor}}\nolimits _ K(L, \overline{K}) = \{ \sigma _1, \ldots , \sigma _ n\} $. Let $M' \subset \overline{K}$ be the $K$-subalgebra generated by $\sigma _ i(L)$, $i = 1, \ldots , n$. It follows from Lemma 9.15.5 that $M'$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma _1(L)$. By uniqueness of normal closure we have $M \cong M'$. Finally, there is a surjective map

\[ L \otimes _ K L \otimes _ K \ldots \otimes _ K L \longrightarrow M', \quad \lambda _1 \otimes \ldots \otimes \lambda _ n \longmapsto \sigma _1(\lambda _1) \ldots \sigma _ n(\lambda _ n) \]

and note that $n \leq [L : K]$ by definition. $\square$

Comments (1)

Comment #9016 by Zhenhua Wu on April 18, 2024 at 03:41

The part ''It follows from Lemma 09HQ that $M′$ is normal over $K$ and that it is the smallest normal subextension of $\overline{K}$ containing $\sigma_i(L)$ needs clarification. To be specific: 1)why is $M'$ an field; 2)how do we use lemma 09HQ to show $M'/K$ is normal; 3) how do we show that it is the smallest normal subextension of $\overline{K}$ containing $\sigma_i(L)$ .

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2 comment(s) on Section 9.16: Splitting fields

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