The Stacks project

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. $\square$

Proof. As $X$ is integral the restriction mapping $\mathcal{O}(X) \to \mathcal{O}(U)$ is injective for any nonempty open subscheme $U \subset X$. Hence we may assume that $X$ is affine. Choose a closed immersion $X \to \mathbf{A}^ n_ k$ and denote $\overline{X}$ the closure of $X$ in $\mathbf{P}^ n_ k$ via the usual immersion $\mathbf{A}^ n_ k \to \mathbf{P}^ n_ k$. Thus we may assume that $X$ is an affine open of a projective variety $\overline{X}$.

Let $\nu : \overline{X}^\nu \to \overline{X}$ be the normalization morphism, see Morphisms, Definition 29.54.1. We know that $\nu $ is finite, dominant, and that $\overline{X}^\nu $ is a normal irreducible scheme, see Morphisms, Lemmas 29.54.5, 29.54.10, and 29.18.2. It follows that $\overline{X}^\nu $ is a proper variety, because $\overline{X} \to \mathop{\mathrm{Spec}}(k)$ is proper as a composition of a finite and a proper morphism (see results in Morphisms, Sections 29.41 and 29.44). It also follows that $\nu $ is a surjective morphism, because the image of $\nu $ is closed and contains the generic point of $\overline{X}$. Hence setting $X^\nu = \nu ^{-1}(X)$ we see that it suffices to prove the result for $X^\nu $. In other words, we may assume that $X$ is a nonempty open of a normal proper variety $\overline{X}$. This case is handled by Lemma 33.28.1. $\square$

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

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

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

Proof. Let $k' \subset \Gamma (X, \mathcal{O}_ X)$ be the integral closure of $k$. Then $X \to \mathop{\mathrm{Spec}}(k)$ factors through $\mathop{\mathrm{Spec}}(k')$, see Schemes, Lemma 26.6.4. As $X$ is reduced we see that $k'$ has no nonzero nilpotent elements. As $k \to k'$ is integral we see that every prime ideal of $k'$ is both a maximal ideal and a minimal prime, and $\mathop{\mathrm{Spec}}(k')$ is totally disconnected, see Algebra, Lemmas 10.36.20 and 10.26.5. As $X$ is connected the morphism $X \to \mathop{\mathrm{Spec}}(k')$ is constant, say with image the point corresponding to $\mathfrak p \subset k'$. Then any $f \in k'$, $f \not\in \mathfrak p$ maps to an invertible element of $\mathcal{O}_ X$. By definition of $k'$ this then forces $f$ to be a unit of $k'$. Hence we see that $k'$ is local with maximal ideal $\mathfrak p$, see Algebra, Lemma 10.18.3. Since we've already seen that $k'$ is reduced this implies that $k'$ is a field, see Algebra, Lemma 10.25.1. $\square$

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). $\square$

Proof. We see that (1) is enough by Proposition 33.28.5. We omit the verification that each of (2), (3), (4) implies (1). $\square$