The Stacks project

Proposition 33.28.5. Let $k$ be a field. Let $X$ be a scheme over $k$. Assume that $X$ is locally of finite type over $k$, connected, reduced, and has finitely many irreducible components. Then $\mathcal{O}(X)^*/k^*$ is a finitely generated abelian group if in addition to the conditions above at least one of the following conditions is satisfied:

  1. the integral closure of $k$ in $\Gamma (X, \mathcal{O}_ X)$ is $k$,

  2. $X$ has a $k$-rational point, or

  3. $X$ is geometrically integral.

Proof. Let $\overline{k}$ be an algebraic closure of $k$. Let $Y$ be a connected component of $(X_{\overline{k}})_{red}$. Note that the canonical morphism $p : Y \to X$ is open (by Morphisms, Lemma 29.23.4) and closed (by Morphisms, Lemma 29.44.7). Hence $p(Y) = X$ as $X$ was assumed connected. In particular, as $X$ is reduced this implies $\mathcal{O}(X) \subset \mathcal{O}(Y)$. By Lemma 33.8.14 we see that $Y$ has finitely many irreducible components. Thus Lemma 33.28.3 applies to $Y$. This implies that if $\mathcal{O}(X)^*/k^*$ is not a finitely generated abelian group, then there exist elements $f \in \mathcal{O}(X)$, $f \not\in k$ which map to an element of $\overline{k}$ via the map $\mathcal{O}(X) \to \mathcal{O}(Y)$. In this case $f$ is algebraic over $k$, hence integral over $k$. Thus, if condition (1) holds, then this cannot happen. To finish the proof we show that conditions (2) and (3) imply (1).

Let $k \subset k' \subset \Gamma (X, \mathcal{O}_ X)$ be the integral closure of $k$ in $\Gamma (X, \mathcal{O}_ X)$. By Lemma 33.28.4 we see that $k'$ is a field. If $e : \mathop{\mathrm{Spec}}(k) \to X$ is a $k$-rational point, then $e^\sharp : \Gamma (X, \mathcal{O}_ X) \to k$ is a section to the inclusion map $k \to \Gamma (X, \mathcal{O}_ X)$. In particular the restriction of $e^\sharp $ to $k'$ is a field map $k' \to k$ over $k$, which clearly shows that (2) implies (1).

If the integral closure $k'$ of $k$ in $\Gamma (X, \mathcal{O}_ X)$ is not trivial, then we see that $X$ is either not geometrically connected (if $k'/k$ is not purely inseparable) or that $X$ is not geometrically reduced (if $k'/k$ is nontrivial purely inseparable). Details omitted. Hence (3) implies (1). $\square$

Comments (0)

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Your email address will not be published.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 04L7. Beware of the difference between the letter 'O' and the digit '0'.