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This might be a stupid question, but I don't understand an easy fact. Let $(X,\mathcal O_X)$ a ringed space.

We know that every module $M$ over a ring $R$ has a free presentation, so why isn't every $\mathcal O_X$-module quasi-coherent?

Why doesn't the free presentation: $\mathcal O_X^{(J)}|_U\rightarrow\mathcal O_X^{(I)}|_U\rightarrow\mathcal F|_U\rightarrow 0$ exist in general?

Many thanks in advance.

Dubious
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    A counterexample is given here: http://math.stackexchange.com/questions/642262/is-skyscraper-sheaf-quasi-coherent – Qiaochu Yuan Feb 23 '16 at 17:49
  • Ok, counterexamples exist. But I'd like to know what goes wrong in general – Dubious Feb 23 '16 at 17:54
  • Maybe the problem is that the indexes $I$ and $J$ should be fixed for the whole open set $U$. – Dubious Feb 23 '16 at 17:57
  • The problem is that locally on $U = Spec (A)$, $\mathcal {F}|{U}$ is not a sheaf given by a module $M$ over $A$. Certainly $\mathcal{F} (U)$ is a module over $A$, but $\mathcal {F}|{U} (D(f)) \cong M_f$ may not hold – user40276 Feb 29 '16 at 11:13

3 Answers3

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1) You write "We know that every module $M$ over a ring $R$ has a free presentation, so why isn't every $\mathcal O_X$-module quasi-coherent? "
But this is begging the question: the correspondence on affine schemes between sheaves and their modules of sections is only valid for quasi-coherent sheaves!

2) Given an arbitrary $\mathcal O_X$- module and a point $x\in X$, there need not even exist a neighbourhood $U$ of $x$ and a surjective morphism of sheaves $\mathcal O_X^{(I)}|U\rightarrow\mathcal F|U\rightarrow 0$.
Indeed, from the Stacks Project (section 17.8) we can extract the following
Example:
Let $X=\mathbb R$ with the usual topology, endowed with the constant sheaf $\mathcal O_X=\underline{\mathbb Z}$ to make it a ringed space.
Let $U =\mathbb R^*_+\subset X$ be the open subspace of positive numbers and let $\mathcal Z=\mathcal O_X\mid U$ be the constant sheaf associated to $\mathbb Z$ on $U$.
Now if $i:U\hookrightarrow \mathbb R$ is the inclusion, consider the sheaf $\mathcal F:=i_!\mathcal Z$ on $X$.
For any connected neighbourhood $U$ of $0$ in X we have $\Gamma(U,\mathcal F)=0$ so that there can be no surjection $\mathcal O _X^{(I)}\mid U\to \mathcal F\mid U\to 0$.
[Recall that morphisms $\mathcal O _X^{(I)}\mid U\to \mathcal F\mid U$ correspond to families $(s_i)$ of sections $s_i\in \Gamma(U,\mathcal F)$]

Georges Elencwajg
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You can always get an exact sequence $ \mathcal{O}_X(U)^I\to \mathcal{O}_X(U)^J\to \mathcal{F}(U)\to 0$, but it is not guaranteed that you can get it to behave well with respect to the restriction maps.

The definition of quasi-coherence you gave is equivalent to the following (which is really helpful to understand): Let $U=Spec(A)$ be an affine open subset of $X$, and let $f\in A$ be a function on $U$. Then the restriction map $\mathcal{F}(U)\to \mathcal{F}(U_f)$ is exactly the localisation of $\mathcal{F}(U)$ with respect to $f$.

Now you should see that this is a non-trivial requirement, and does not follow from the sheaf axiom.

edo arad
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Indeed, every $R$-module has a free presentation; and the proof of this fact is constructive, hence also applies in sheaf toposes (by their "internal language"). Run there, it shows that every sheaf $\mathcal{F}$ of $\mathcal{O}_X$-modules has a free presentation $$ \mathcal{O}_X\langle \mathcal{A} \rangle \to \mathcal{O}_X\langle \mathcal{B} \rangle \to \mathcal{O}_X\langle \mathcal{F} \rangle \to 0, $$ where $\mathcal{A}$ and $\mathcal{B}$ are certain sheaves of sets.

But the condition for a sheaf of $\mathcal{O}_X$-modules to be quasicoherent is much stronger: For that, there needs to be a presentation as above where $\mathcal{A}$ and $\mathcal{B}$ are locally constant.

Ingo Blechschmidt
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