Lemma 33.28.3. Let $k$ be an algebraically closed field. Let $X$ be a connected reduced scheme which is locally of finite type over $k$ with finitely many irreducible components. Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.

Proof. Let $X = \bigcup X_ i$ be the irreducible components. By Lemma 33.28.2 we see that $\mathcal{O}(X_ i)^*/k^*$ is a finitely generated abelian group. Let $f \in \mathcal{O}(X)^*$ be in the kernel of the map

$\mathcal{O}(X)^* \longrightarrow \prod \mathcal{O}(X_ i)^*/k^*.$

Then for each $i$ there exists an element $\lambda _ i \in k$ such that $f|_{X_ i} = \lambda _ i$. By restricting to $X_ i \cap X_ j$ we conclude that $\lambda _ i = \lambda _ j$ if $X_ i \cap X_ j \not= \emptyset$. Since $X$ is connected we conclude that all $\lambda _ i$ agree and hence that $f \in k^*$. This proves that

$\mathcal{O}(X)^*/k^* \subset \prod \mathcal{O}(X_ i)^*/k^*$

and the lemma follows as on the right we have a product of finitely many finitely generated abelian groups. $\square$

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