Is the mapping from a scheme to its global sections a closed map? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:37:56Z http://mathoverflow.net/feeds/question/13506 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13506/is-the-mapping-from-a-scheme-to-its-global-sections-a-closed-map Is the mapping from a scheme to its global sections a closed map? Anweshi 2010-01-30T21:12:38Z 2010-02-01T03:22:25Z <p>This is a question posed to me in private communication by <a href="http://mathoverflow.net/users/3582/amaanush" rel="nofollow">this user</a>.</p> <p>Given a scheme $T$, let $\Gamma (T) = Mor (T, \mathbb{A}^1)$ be the ring of global sections. Note that there is a canonical map $\phi : T \rightarrow Spec (\Gamma (T))$.</p> <p>Is $\phi$ a closed mapping onto the image, ie. is $im\ (Z)$ a closed subset of $im(T)$ for all closed subsets $Z$ of $T$ ?.</p> http://mathoverflow.net/questions/13506/is-the-mapping-from-a-scheme-to-its-global-sections-a-closed-map/13509#13509 Answer by Martin Brandenburg for Is the mapping from a scheme to its global sections a closed map? Martin Brandenburg 2010-01-30T21:28:23Z 2010-01-30T21:35:03Z <p>Note that if $T$ is reduced, this morphism has dense image.</p> <p>If $T$ is a locally compact totally disconnected hausdorff space, it can be given explicitely a reduced scheme structure, which is affine iff $T$ is compact. Besides $T \to Spec \Gamma(T)$ is an dense open immersion. Thus it is not closed as long $T$ is not compact. I've proven this <a href="http://maddin.110mb.com/pdf/hausdorff%5Fneu.pdf" rel="nofollow">here</a>. Thus a counterexample would be $T = \mathbb{Q}_p$.</p> <p>I expect that others will show you more "natural" examples.</p> http://mathoverflow.net/questions/13506/is-the-mapping-from-a-scheme-to-its-global-sections-a-closed-map/13515#13515 Answer by Qing Liu for Is the mapping from a scheme to its global sections a closed map? Qing Liu 2010-01-30T22:01:27Z 2010-01-30T22:01:27Z <p>Here is a ''natural'' example as expected by Martin. Let $T$ be the projective line over ${\mathbb Z}$, minus a rational point $x_0$ of the closed fiber at some prime $p$. Then $O(T)=\mathbb Z$ (direct computation or use Zariski's extension theorem for normal schemes), $\phi$ is just the structural morphism and is onto. Let $Z'$ be a section of the projective line passing through $x_0$. Then $Z=Z'\cap T$ is closed in $T$, but $\phi(Z)$ is not closed in $\phi(T)$ because it is the image of the composition $Z\to {\rm Spec} O(Z)\to {\rm Spec} O(T)$, and as $Z$ is affine, the image is just the complementary of the closed point $p$. </p> http://mathoverflow.net/questions/13506/is-the-mapping-from-a-scheme-to-its-global-sections-a-closed-map/13520#13520 Answer by Emerton for Is the mapping from a scheme to its global sections a closed map? Emerton 2010-01-31T00:15:38Z 2010-02-01T03:22:25Z <p>[Added: I misread the question, and in fact this answer does not answer the OP's question, but rather the following question: is $\phi(T)$ closed in Spec $\Gamma(T)$, which is a different question. Probably the upvotes can be attributed to the link to the stacks project!]</p> <p>If $T$ is quasi-affine (i.e. admits an open immersion into affine space), then the map $\phi$ is an open immersion, and in fact Spec $\Gamma(T)$ is the initial object in the category of affine schemes containing $T$ as an open subscheme. </p> <p>In particular, in this case $\phi$ has closed image if and only if and only if $T$ is in fact affine. [Added: As Qing Liu points out in a comment below, in this quasi-affine situation, $\phi$ is in fact a closed map onto its image.]</p> <p>Thus if we take $T$ to be ${\mathbb A}^2_k \setminus \{0\}$ for some field $k$, i.e. affine $2$-space with the origin removed, then we get an example of $T$ where this map is open with non-closed image (since this $T$ is quasi-affine but not affine). Note that Spec $\Gamma(T) = {\mathbb A^2}_k$.</p> <p>(This is a geometric analogue of Qing Lui's more arithmetic example; what both have in common is that a closed point was removed from a 2-dimensional affine scheme, so as to make a quasi-affine scheme that is not affine.[Added: I also misread Qing Liu's example; my remark would apply to the affine line over ${\mathbb Z}$ with a closed point removed; Qing's example is more complicated, since it is actually dealing with the OP's question. One can make a geometric analogue of Qing's example by deleting a closed point from ${\mathbb A}^1\times {\mathbb P}^1$; more geometrically still, remove one of the lines of a ruling from a projective quadric surface, and then remove an additional point.])</p> <p>EDIT: In the definition of quasi-affine, one should also require that $T$ be quasi-compact. (The <a href="http://www.math.columbia.edu/algebraic_geometry/stacks-git/" rel="nofollow">stacks project</a> is a terrific resource for these foundational definitions in scheme theory, particularly with regard to finiteness and separation issues.)</p> <p>Note that if $T$ is any quasi-compact scheme, then the map $T \to$ Spec $\Gamma(T)$ has dense image. (If $f \in \Gamma(T)$ and $D(f)$ is the usual affine open in Spec $\Gamma(T)$, i.e. Spec $\Gamma(T)_f$, then if $\phi^{-1}(D(f))$ is empty, it must be that $f$ is locally nilpotent on $T$. Since $T$ is quasi-compact this implies that $f$ is actually nilpotent, and hence that $D(f)$ is empty.) As Martin notes in his answer, this is similarly true if $T$ is reduced.</p> <p>It need not be true if $T$ is non-reduced and non-quasi-compact (since $T$ may then admit locally nilpotent sections of $\mathcal O_T$ that are not globally nilpotent, e.g. $T = \coprod_n$ Spec $k[x]/(x^n)$).</p>