The Stacks project

33.28 Groups of invertible functions

It is often (but not always) the case that $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group if $X$ is a variety over $k$. We show this by a series of lemmas. Everything rests on the following special case.

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 29.42.11) and compositions of proper morphisms are proper (Morphisms, Lemma 29.39.4). 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 10.30.1, Properties, Definition 28.3.1, Properties, Lemmas 28.5.2 and 28.7.2, Morphisms, Lemma 29.14.2.

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 29.14.6, 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 10.118.7. Hence we obtain a valuation

\[ v_ j : k(\overline{X})^* \longrightarrow \mathbf{Z} \]

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

\[ \mathcal{O}(X)^* \longrightarrow \prod \nolimits _{j \in J} \mathbf{Z}, \quad f \longmapsto \prod v_ j(f) \]

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 10.152.6. 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. $\square$

Next, we generalize the case above by some elementary arguments, still keeping the field algebraically closed.

Lemma 33.28.2. Let $k$ be an algebraically closed field. Let $X$ be an integral scheme locally of finite type over $k$. Then $\mathcal{O}^*(X)/k^*$ is a finitely generated abelian group.

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 29.52.1. We know that $\nu $ is finite, dominant, and that $\overline{X}^\nu $ is a normal irreducible scheme, see Morphisms, Lemmas 29.52.5, 29.52.9, and 29.17.2. 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 29.39 and 29.42). 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. $\square$

The preceding lemma implies the following slight generalization.

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. $\square$

Lemma 33.28.4. Let $k$ be a field. Let $X$ be a scheme over $k$ which is connected and reduced. Then the integral closure of $k$ in $\Gamma (X, \mathcal{O}_ X)$ is a field.

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 10.35.20 and 10.25.5. 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 10.17.2. Since we've already seen that $k'$ is reduced this implies that $k'$ is a field, see Algebra, Lemma 10.24.1. $\square$

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 29.22.4) and closed (by Morphisms, Lemma 29.42.7). 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 33.8.13 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 $k \subset k'$ is not purely inseparable) or that $X$ is not geometrically reduced (if $k \subset k'$ is nontrivial purely inseparable). Details omitted. Hence (3) implies (1). $\square$

Lemma 33.28.6. Let $k$ be a field. Let $X$ be a variety over $k$. The group $\mathcal{O}(X)^*/k^*$ is a finitely generated abelian group provided at least one of the following conditions holds:

  1. $k$ is integrally closed in $\Gamma (X, \mathcal{O}_ X)$,

  2. $k$ is algebraically closed in $k(X)$,

  3. $X$ is geometrically integral over $k$, or

  4. $k$ is the “intersection” of the field extensions $k \subset \kappa (x)$ where $x$ runs over the closed points of $x$.

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


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 04L3. Beware of the difference between the letter 'O' and the digit '0'.