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:

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

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

$X$ 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).
$\square$

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