Lemma 33.28.1. Let $k$ be an algebraically closed field. Let $\overline{X}$ be a proper variety over $k$. Let $X \subset \overline{X}$ be an open subscheme. Assume $X$ is normal. Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.

**Proof.**
Since the statement only concerns $X$, we may replace $\overline{X}$ by a different proper variety over $k$. Let $\nu : \overline{X}^\nu \to \overline{X}$ be the normalization morphism. By Lemma 33.27.1 we have that $\nu $ is finite and $\overline{X}^\nu $ is a variety. Since $X$ is normal, we see that $\nu ^{-1}(X) \to X$ is an isomorphism (tiny detail omitted). Finally, we see that $\overline{X}^\nu $ is proper over $k$ as a finite morphism is proper (Morphisms, Lemma 29.44.11) and compositions of proper morphisms are proper (Morphisms, Lemma 29.41.4). Thus we may and do assume $\overline{X}$ is normal.

We will use without further mention that for any affine open $U$ of $\overline{X}$ the ring $\mathcal{O}(U)$ is a finitely generated $k$-algebra, which is Noetherian, a domain and normal, see Algebra, Lemma 10.31.1, Properties, Definition 28.3.1, Properties, Lemmas 28.5.2 and 28.7.2, Morphisms, Lemma 29.15.2.

Let $\xi _1, \ldots , \xi _ r$ be the generic points of the complement of $X$ in $\overline{X}$. There are finitely many since $\overline{X}$ has a Noetherian underlying topological space (see Morphisms, Lemma 29.15.6, Properties, Lemma 28.5.5, and Topology, Lemma 5.9.2). For each $i$ the local ring $\mathcal{O}_ i = \mathcal{O}_{X, \xi _ i}$ is a normal Noetherian local domain (as a localization of a Noetherian normal domain). Let $J \subset \{ 1, \ldots , r\} $ be the set of indices $i$ such that $\dim (\mathcal{O}_ i) = 1$. For $j \in J$ the local ring $\mathcal{O}_ j$ is a discrete valuation ring, see Algebra, Lemma 10.119.7. Hence we obtain a valuation

with the property that $v_ j(f) \geq 0 \Leftrightarrow f \in \mathcal{O}_ j$.

Think of $\mathcal{O}(X)$ as a sub $k$-algebra of $k(X) = k(\overline{X})$. We claim that the kernel of the map

is $k^*$. It is clear that this claim proves the lemma. Namely, suppose that $f \in \mathcal{O}(X)$ is an element of the kernel. Let $U = \mathop{\mathrm{Spec}}(B) \subset \overline{X}$ be any affine open. Then $B$ is a Noetherian normal domain. For every height one prime $\mathfrak q \subset B$ with corresponding point $\xi \in X$ we see that either $\xi = \xi _ j$ for some $j \in J$ or that $\xi \in X$. The reason is that $\text{codim}(\overline{\{ \xi \} }, \overline{X}) = 1$ by Properties, Lemma 28.10.3 and hence if $\xi \in \overline{X} \setminus X$ it must be a generic point of $\overline{X} \setminus X$, hence equal to some $\xi _ j$, $j \in J$. We conclude that $f \in \mathcal{O}_{X, \xi } = B_{\mathfrak q}$ in either case as $f$ is in the kernel of the map. Thus $f \in \bigcap _{\text{ht}(\mathfrak q) = 1} B_{\mathfrak q} = B$, see Algebra, Lemma 10.157.6. In other words, we see that $f \in \Gamma (\overline{X}, \mathcal{O}_{\overline{X}})$. But since $k$ is algebraically closed we conclude that $f \in k$ by Lemma 33.26.2. $\square$

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