The Stacks project

Lemma 9.15.7. Let $L/M/K$ be a tower of algebraic extensions.

  1. If $M/K$ is normal, then any automorphism $\tau $ of $L/K$ induces an automorphism $\tau |_ M : M \to M$.

  2. If $L/K$ is normal, then any $K$-algebra map $\sigma : M \to L$ extends to an automorphism of $L$.

Proof. Choose an algebraic closure $\overline{L}$ of $L$ (Theorem 9.10.4).

Let $\tau $ be as in (1). Then $\tau (M) = M$ as subfields of $\overline{L}$ by Lemma 9.15.5 and hence $\tau |_ M : M \to M$ is an automorphism.

Let $\sigma : M \to L$ be as in (2). By Lemma 9.10.5 we can extend $\sigma $ to a map $\tau : L \to \overline{L}$, i.e., such that

\[ \xymatrix{ L \ar[r]_\tau & \overline{L} \\ M \ar[u] \ar[ru]_\sigma & K \ar[l] \ar[u] } \]

is commutative. By Lemma 9.15.5 we see that $\tau (L) = L$. Hence $\tau : L \to L$ is an automorphism which extends $\sigma $. $\square$

Comments (2)

Comment #4340 by Yairon Cid Ruiz on

Small typo in part (2): It should be added "any".

There are also:

  • 3 comment(s) on Section 9.15: Normal extensions

Post a comment

Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BME. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 9.15.7 as pdf