Lemma 9.14.8. Let $K/F$ be a finite extension. Let $\overline{F}$ be an algebraic closure of $F$. Then $[K : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|$.

**Proof.**
We first prove this when $K/F$ is purely inseparable. Namely, we claim that in this case there is a unique map $K \to \overline{F}$. This can be seen by choosing a sequence of elements $\alpha _1, \ldots , \alpha _ n \in K$ as in Lemma 9.14.5. The irreducible polynomial of $\alpha _ i$ over $F(\alpha _1, \ldots , \alpha _{i - 1})$ is $x^ p - \alpha _ i^ p$. Applying Lemma 9.12.9 we see that $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = 1$. On the other hand, $[K : F]_ s = 1$ in this case hence the equality holds.

Let's return to a general finite extension $K/F$. In this case choose $F \subset K_ s \subset K$ as in Lemma 9.14.6. By Lemma 9.12.11 we have $|\mathop{\mathrm{Mor}}\nolimits _ F(K_ s, \overline{F})| = [K_ s : F] = [K : F]_ s$. On the other hand, every field map $\sigma ' : K_ s \to \overline{F}$ extends to a unique field map $\sigma : K \to \overline{F}$ by the result of the previous paragraph. In other words $|\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})| = |\mathop{\mathrm{Mor}}\nolimits _ F(K_ s, \overline{F})|$ and the proof is done. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: