MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $T$ ?.

share|cite|improve this question
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 $p$.

share|cite|improve this answer
Easier geometric version: $T=(\mathbb A^1\times\mathbb P^1)\setminus 0$. – VA. Feb 1 '10 at 3:52

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

share|cite|improve this answer
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.