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
of $K$-algebras where the number of tensors can be taken $[L : K]_ s \leq [L : K]$.
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
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 $\sigma _ i(L)$, $i = 1, \ldots , n$. Then $M'$ is a field since any $K$-subalgebra of $\overline{K}$ is a field. Any $K$-algebra map from $M'$ to $\overline{K}$ permutes the $\sigma _ i$ so sends $M'$ into and onto $M'$. By construction the field $M'$ is generated by conjugates of elements of $\sigma _1(L)$. Having said this 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
and note that $n \leq [L : K]$ by definition. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #9016 by Zhenhua Wu on
Comment #9195 by Stacks project on
There are also: