Let $X$ be a noetherian scheme separated and of finite type over a noetherian base $S$ and $\epsilon$ a section of its structure morphism. I want to consider $S$ via this section as a closed subscheme of $X$.

Given a coherent sheaf $F$ on $X$ with the property $supp(F)=S$ (here consider $S$ embedded in $X$), where $supp$ denotes the support of a sheaf, then my question is:

Does one find a surjective module homomorphism

$F \rightarrow \epsilon_*\mathcal O_S$?

It does not need to be canonical; e.g. if $S$ is the spectrum of a field $k$ one certainly does, by Adjunction and simply chosing a projection of the $k-$vector space $F/\mathcal m$ to $k$, where $\mathcal m$ is the maximal ideal of the local ring of $\epsilon(pt)$, with $pt$ the point of $Spec(k)$.

I thought a pretty while about it, but don't see how one would get the epimorphism in the general case in question, so my guess would be no, in general. But maybe some additional properties I don't see at the moment would improve the situation.


When you say that $supp(F) = S$, do you mean this scheme-theoretically, i.e. does the action of the structure sheaf $\mathcal O_X$ on $F$ factor through $\epsilon_*\mathcal O_S$?

Assuming that the answer is yes (and even if the answer is no, what I am describing is still subcase of the more general situation), you can ignore $X$ and just work on $S$. Now you are asking whether there is a surjection $F \to \mathcal O_S$. Let's suppose $S$ is affine (again, a special case of your situation): then we can pass to modules, and you are asking if, given a Noetherian ring $A$ and a finitely generated and faithful $A$-module $M$ (faithful is here because $M$ has full support by assumption), there is a surjection $M \to A$.

Clearly the answer is no in general. (E.g. let $A = k[x,y]$ and let $M$ be the ideal $(x,y)$.) On the other hand the answer is yes is $A$ is a Dedekind domain (because then torsion free implies free), or if $M$ is free.

In the general (i.e. not necessarily affine) case, suppose that $F$ is locally free. Then giving a surjection $F \to \mathcal O_S$ is the same as giving an injection $\mathcal O_S \to F^{\vee}$ whose cokernel is again locally free. If $F$ is the sheaf of sections of the vector bundle $\mathcal V$, then this is equivalent to asking that $\mathcal V^{\vee}$, admits the trivial line bundle as a subbundle, or that $\mathcal V$ admits the trivial bundle as a quotient bundle.

Added: Let $\mathcal I$ denote the ideal sheaf of $S$ (thought of as a closed subscheme of $X$ via $\epsilon$). If there exists a map $F \to \epsilon_* \mathcal O_S$, then it factors through $F/\mathcal I F$. Thus it is no loss of generality to consider to restrict to the situation where $F$ is scheme-theoretically supported on $S$, i.e. where $\mathcal I$ annihilates $F$. This is the case described above, and as remarked there, the answer is no in general.

  • $\begingroup$ When I say $supp(F)=S$, then I just mean: the points on $X$, where the stalk of $F$ is nonzero, are precisely the points in $S$. And I'm not sure if I can ignore $X$, because from $supp(F)=S$ it only follows that $F$ is killed by a power of the coherent defining ideal of $S$, i.e. $F$ is only a module on some infinitesimal neighborhood of $S$ in $X$. $\endgroup$ – Cyril Jan 5 '12 at 20:36
  • $\begingroup$ @Cyril: Dear Cyril, Then, as I wrote, the case where you can ignore $X$ is a special case, in which nevertheless the answer is no in general. Actually, after a moment's reflection, I added a remark about the more general case, explaining why it reduces to the special case I consider. Regards, $\endgroup$ – Matt E Jan 5 '12 at 20:50
  • $\begingroup$ Now it became clear to me, thanks! $\endgroup$ – Cyril Jan 5 '12 at 21:11
  • $\begingroup$ @Cyril: Dear Cyril, You're welcome. Also, I correct an erroneous assertion. If $F$ is locally free on $S$, then your condition is equivalent to asking that the corresponding vector bundle $\mathcal V$ admit the trivial bundle as a quotient bundle (which is weaker than asking that it admit the trivial bundle as a summand). Regards, $\endgroup$ – Matt E Jan 5 '12 at 21:31

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