Lemma 10.34.2. Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is of finite type over $R$, then there exists an $f \in R$ such that $R_ f$ is a field, and $R_ f \subset K$ is a finite field extension.

**Proof.**
By Lemma 10.30.2 there exist a nonempty open $U \subset \mathop{\mathrm{Spec}}(R)$ contained in the image $\{ (0)\} $ of $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(R)$. Choose $f \in R$, $f \not= 0$ such that $D(f) \subset U$, i.e., $D(f) = \{ (0)\} $. Then $R_ f$ is a domain whose spectrum has exactly one point and $R_ f$ is a field. Then $K$ is a finitely generated algebra over the field $R_ f$ and hence a finite field extension of $R_ f$ by the Hilbert Nullstellensatz (Theorem 10.34.1).
$\square$

## 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Your email address will not be published.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)