Lemma 9.8.5. A finite extension is algebraic. In fact, an extension $E/k$ is algebraic if and only if every subextension $k(\alpha )/k$ generated by some $\alpha \in E$ is finite.

**Proof.**
Let $E/k$ be finite, say of degree $n$. Choose $\alpha \in E$. Then the elements $\{ 1, \alpha , \ldots , \alpha ^ n\} $ are linearly dependent over $k$, or we would necessarily have $[E : k] > n$. A relation of linear dependence now gives the desired polynomial that $\alpha $ must satisfy.

For the last assertion, note that a monogenic extension $k(\alpha )/k$ is finite if and only if $\alpha $ is algebraic over $k$, by Examples 9.7.4 and 9.7.6. So if $E/k$ is algebraic, then each $k(\alpha )/k$, $\alpha \in E$, is a finite extension, and conversely. $\square$

## 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Your email address will not be published.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: