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This is a question posed to me in private communication by this user.

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 $\phi : T \rightarrow Spec (\Gamma (T))$.

Is $\phi$ a closed mapping onto the image, ie. is $im\ (Z)$ a closed subset of $im(T)$ for all closed subsets $Z$ of $T$ ?.

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3 Answers 3

up vote 9 down vote accepted

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 $p$.

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Easier geometric version: $T=(\mathbb A^1\times\mathbb P^1)\setminus 0$. –  VA. Feb 1 '10 at 3:52
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[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!]

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.

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.]

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 $\Gamma(T) = {\mathbb A^2}_k$.

(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.])

EDIT: In the definition of quasi-affine, one should also require that $T$ be quasi-compact. (The stacks project is a terrific resource for these foundational definitions in scheme theory, particularly with regard to finiteness and separation issues.)

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.

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 $k[x]/(x^n)$).

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Dear Matt, thanks for the wonderful link to the stacks projet ! I didn't know the open immersion for quasi-affine schemes. I also thought to the punctured affine plane, but the OP asks whether $\phi(Z)$ is closed in $\phi(T)$, which is OK when $\phi$ is an immersion. So I had to complicated a little the example. But probably it is not so interesting to consider $\phi(Z)$ inside $\phi(T)$. –  Qing Liu Jan 31 '10 at 21:22
    
Dear Qing, Oh dear, I misread the question. As you could probably tell, I was answering the question as to whether $\phi(T)$ was closed in Spec $\Gamma(T)$, rather than the question as to whether $\phi$ was a closed mapping onto its image. I will edit my answer to remark on this, and to corret my description of your example. –  Emerton Feb 1 '10 at 2:01
    
I had in fact this question also in mind .. I was going to that question next. Saves me the trouble. Thanks a lot .. However I must accept Q. Liu's answer instead. I keep doing this to your answers though they are very good; I hope you don't mind. –  Anweshi Feb 5 '10 at 12:26
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Note that if $T$ is reduced, this morphism has dense image.

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 here. Thus a counterexample would be $T = \mathbb{Q}_p$.

I expect that others will show you more "natural" examples.

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