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

Here is something that isn't yet very clear to me. Say, I've got a commutative ring A. I consider the affine scheme from A, so it's a sheaf of rings over Spec A.

EDIT: And additionally let's say Spec A is Hausdorff.

Now additionally let's say I know an A-module M and from that I can make a sheave of modules over O_Spec(A), call it M~. All standard stuffs till now. But now I want to have one more information, namely I have an open subset of Spec(A), say U, that is dense in Spec(A). And I know additionally that the stalks M~_x are isomorphic to O_x for all x in U.. Can one conclude that M and A are A-module isomorphic? (if so can one follow the same argument for general schemes with modules over them?) What are the conditions by which one can conclude this?

share|cite|improve this question
I just edited the question to be specific on Spec A being Hausdorff – Jose Capco Nov 18 '09 at 21:08
The Hausdorffness assumption changes the question from "reasonable" to "rather pathological". Do you have in mind an example of a Hausdorff affine scheme Spec(A) and a dense open subset U for which U isn't all of Spec(A)? Maybe if you explained your example I'd be able to make more sense of the question in its current form. – Kevin Buzzard Nov 18 '09 at 21:22
If you take any Stone space with infinite number of non-isolated point (say the spectrum of infinite product of fields) and remove any finite such non-isolated point from it, then you get an open subset U of that space that is dense in it. That such a Stone space is the spectrum of a ring roughly comes from Stone's representation theorem. – Jose Capco Nov 18 '09 at 22:18
The answer to the question (as it currently stands) is still "no", I believe, because M can be A+A/P for P a prime ideal corresponding to a point not in U (or A+(A/P)^S for a set S of cardinality bigger than A, if you really want to make sure it's not isomorphic to A...). By the way, it seems to me to be a bad idea to change a question after you've asked it, without writing EDIT in big letters where you changed it. It seems to make things much more confusing especially if people have already left answers. – Kevin Buzzard Nov 18 '09 at 22:37
Ok.. Ill add "EDIT". I did wrote and meant Haussdorf in the original question, except that I wrote "let Spec A have a comfortable topology, say Haussdorf" and i thought that was rather confusing, so I wrote just "let Spec A be Haussdorf". Thanks. – Jose Capco Nov 18 '09 at 22:42
up vote 2 down vote accepted

If U is an open set in X, but U isn't X, then there are non-zero sheaves on X whose support lies outside U. Now add O_X to one of these to get a counterexample.

share|cite|improve this answer
Ok, now I am a bit confused. In the comments of the original post. You write M=A+A/P (for P non-isolated point of Spec A). I assumed here direct sum. Could you briefly explain why the stalks of M_Q is isomorphic to A_Q (which is isomorphic to the field A/Q as Spec A is Hausdroff) for all Q in Spec A\{P} .. I suspect that this is not true. – Jose Capco Nov 21 '09 at 21:26
I didn't claim what you're asking me to prove above. I only claimed it was true for Q in U, which is open and doesn't contain P, so in particular Q isn't in the closure of {P}. Does this clarify things or do you still think I've slipped up? – Kevin Buzzard Nov 22 '09 at 9:36
no sorry, right. Im alwayss thinking of M as a ring, but I shouldn't really. – Jose Capco Nov 22 '09 at 19:58

The answer is no - the point is that finitely generated projective modules are locally free but not necessarily globally so.

For instance take a Dedekind domain $A$ which does not have unique factorization and consider a non-principal prime ideal $P$. Then $\tilde{P}_x \cong \mathcal{O}_x$ for any $x\in Spec A$ but it is not isomorphic to $A$.

This will occur for any scheme with non-trivial Picard group, in the sense that there will be line bundles (i.e. locally free sheaves of rank 1) which are not trivial.

Edit If you really do want $Spec A$ Hausdorff then off the top of my head one can say the following. If $A$ is noetherian then since it must be dimension 0 it is artinian and so a product of artin local rings. So the only dense subset in the spectrum is the whole thing and so any line bundle is trivial as it is a product of line bundles over local affine schemes.

share|cite|improve this answer
But I assumed Spec A to be Haussdorff.. that's not true for a Dedekind domain – Jose Capco Nov 18 '09 at 20:50
Yes - I just edited it - do you really want Spec A to be Hausdorff? If so a different answer is required. – Greg Stevenson Nov 18 '09 at 20:52
Yes and Yes. I'd understand it for Spec A not being zero dimensional, but it gets intruiging if Spec A is zero dimensional. But I suspect that even for Spec A Hausdorff you get a non-example, but you may easily find necessary and sufficient condition for this to work. – Jose Capco Nov 18 '09 at 21:01

Here's a typical example of $M$ with the property that $O_x = M_x$ for $x$ in open subset.

Take $U = \mathop{\mathrm{Spec}}A-\{f=0\}$. Note that $U$ is $\mathop{\mathrm{Spec}} A_f$ where $A_f$ is a localization of $A$, defined as a ring of fractions of the form $m/f^n$, $m\in A$. Now $A_f$ is an $A$-module, so it gives rise to a quasicoherent sheaf on $\mathop{\mathrm{Spec}}A$ which is the same sheaf as $\mathcal O$ over $U$.

However, you statement should be true if you restrict to coherent sheaves $M$, essentially because the example above is not possible any more. I think this is an exercise in the chapter of Hartshorne that deals with coherent sheaves.

share|cite|improve this answer
No, it shouldn't be true for coherent sheaves. See the counter-examples above. There might be some interesting cases where it is true for coherent + torsion free. But the questioner seems very confused; I don't understand what he's looking for. – David Speyer Nov 19 '09 at 1:39
I think I answered a wrong question. – Ilya Nikokoshev Nov 19 '09 at 6:07
No harm taken :) what quetion was it you thought you answered? – Jose Capco Nov 19 '09 at 9:11
Something about coherent sheaves on smooth schemes I guess... – Ilya Nikokoshev Nov 19 '09 at 20:29

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.