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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?

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  • $\begingroup$ I just edited the question to be specific on Spec A being Hausdorff $\endgroup$
    – Jose Capco
    Commented Nov 18, 2009 at 21:08
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    $\begingroup$ 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. $\endgroup$
    – Kevin Buzzard
    Commented Nov 18, 2009 at 21:22
  • $\begingroup$ 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. $\endgroup$
    – Jose Capco
    Commented Nov 18, 2009 at 22:18
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    $\begingroup$ 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. $\endgroup$
    – Kevin Buzzard
    Commented Nov 18, 2009 at 22:37
  • $\begingroup$ 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. $\endgroup$
    – Jose Capco
    Commented Nov 18, 2009 at 22:42

3 Answers 3

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jose Capco
    Commented Nov 21, 2009 at 21:26
  • $\begingroup$ 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? $\endgroup$
    – Kevin Buzzard
    Commented Nov 22, 2009 at 9:36
  • $\begingroup$ no sorry, right. Im alwayss thinking of M as a ring, but I shouldn't really. $\endgroup$
    – Jose Capco
    Commented Nov 22, 2009 at 19:58
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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.

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  • $\begingroup$ But I assumed Spec A to be Haussdorff.. that's not true for a Dedekind domain $\endgroup$
    – Jose Capco
    Commented Nov 18, 2009 at 20:50
  • $\begingroup$ Yes - I just edited it - do you really want Spec A to be Hausdorff? If so a different answer is required. $\endgroup$
    – Greg Stevenson
    Commented Nov 18, 2009 at 20:52
  • $\begingroup$ 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. $\endgroup$
    – Jose Capco
    Commented Nov 18, 2009 at 21:01
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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.

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  • $\begingroup$ 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. $\endgroup$
    – David E Speyer
    Commented Nov 19, 2009 at 1:39
  • $\begingroup$ I think I answered a wrong question. $\endgroup$
    – Ilya Nikokoshev
    Commented Nov 19, 2009 at 6:07
  • $\begingroup$ No harm taken :) what quetion was it you thought you answered? $\endgroup$
    – Jose Capco
    Commented Nov 19, 2009 at 9:11
  • $\begingroup$ Something about coherent sheaves on smooth schemes I guess... $\endgroup$
    – Ilya Nikokoshev
    Commented Nov 19, 2009 at 20:29

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