Lemma 9.8.10. Let $E/F$ be a finite or more generally an algebraic extension of fields. Any subring $F \subset R \subset E$ is a field.

**Proof.**
Let $\alpha \in R$ be nonzero. Then $1, \alpha , \alpha ^2, \ldots $ are contained in $R$. By Lemma 9.8.5 we find a nontrivial relation $a_0 + a_1 \alpha + \ldots + a_ d \alpha ^ d = 0$. We may assume $a_0 \not= 0$ because if not we can divide the relation by $\alpha $ to decrease $d$. Then we see that

which proves that the inverse of $\alpha $ is the element $a_0^{-1} (- a_1 - \ldots - a_ d \alpha ^{d - 1})$ of $R$. $\square$

## 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Your email address will not be published.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: