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The center of a category $C$ is the monoid $Z(C)=\mathrm{End}_{[C,C]}(\mathrm{id}_C)$. Thus it consists of all families of endomorphisms $M \to M$ of objects $M \in C$, such that for every morphism $M \to N$ the resulting diagram commutes. If $C$ is an $Ab$-category, this actually becomes a ring. For example, the center of $\mathrm{Mod}(A)$ is the center of $A$, if $A$ is a (noncommutative) ring.

Now my question is: What is the center of the category of quasi-coherent modules $\mathrm{Qcoh}(X)$ on a scheme $X$? Observe that there is a natural map $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$, mapping a global section to the endomorphisms of the quasi-coherent modules which are given by multiplication with this section. Also, there is a natural map $Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$, which takes a compatible family of endomorphisms to the image of the global section $1$ in $\mathcal{O}_X$. The composite $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X)) \to \Gamma(X,\mathcal{O}_X)$ is the identity, but what about the other composite? If $X$ is affine, it also turns out to be the identity.

In the end of his thesis about the Reconstruction Theorem, Gabriel proves that $\Gamma(X,\mathcal{O}_X) \to Z(\mathrm{Qcoh}(X))$ is an isomorphism if $X$ is a noetherian scheme (using recollements of localizing subcategories). I'm pretty sure that the proof just uses that $X$ is quasi-compact and quasi-separated. Now what about the general case?

Note that this is about the reconstruction of the structure sheaf $\mathcal{O}_X$ from $X$. Also note that Rosenberg claims in Reconstruction of schemes that this is possible for arbitrary schemes. But if I understand correctly, Rosenberg uses a structure sheaf on the spectrum of an abelian category which avoids the above problems and uses $Z(\mathrm{Mod}(X))=\Gamma(X,\mathcal{O}_X)$, which is certainly true (use extensions by zero), but doesn't yield the result for $\mathrm{Qcoh}(X)$. But I'm not sure because Rosenberg refers to a proof step (a4) which is not there.

Edit: Angelo has proven the result below if $X$ is quasi-separated. Now what happens if $X$ is not quasi-separated? In that case, $\mathrm{Qcoh}(X)$ might have not "enough" objects, which makes the question much harder (and perhaps also less interesting).

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2 Answers 2

In the case of a quasi-separated scheme, the center of the category of quasi-coherent sheaves is $\mathcal O(X)$. Suppose that $f$ is in the center. Let $a \in \mathcal O(X)$ be the scalar that describes the action of $f$ on $\mathcal O_X$; it is enough to show that if $a = 0$ then $f = 0$. Suppose that $M$ is a quasi-coherent sheaf, and that $s$ is a section of $M$ over an open subscheme $U$ of $X$; we need to show that $f_M(s) = 0$. Call $j\colon U \to X$ the embedding; then $j$ is quasi-compact, because $X$ is quasi-separated, so $\overline M := j_*(M\mid_U)$ is quasi-coherent. The adjuntion map $M \to \overline M$ induces an isomophism $M(U) \simeq \overline M(U)$. Call $\overline s$ the image of $s$ in $\overline M$; is enough to show that $f_{\overline M}\overline s = 0$. But $\overline s$ extends to all of $X$, so it in the image of a map $\mathcal O_X \to \overline M$, and the thesis follows.

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There is no such natural map $M \to \overline{M}$, since it is not compatible with respect to the restriction of $X$ to $U$. –  Martin Brandenburg Oct 11 '10 at 15:47
Does this version work? –  Angelo Oct 11 '10 at 16:01
Yes, of course. I wonder why I have not seen this, because I already started with $\overline{M}$ before asking the question here ... –  Martin Brandenburg Oct 11 '10 at 17:05
Being a bit picky, I think in this argument we need U to be quasi-compact. Otherwise I don't see why $$U\rightarrow X$$ is quasi-compact. –  Yuhao Huang Oct 25 '11 at 5:38
@Yuhao Huang: right, but it suffices to show that $f_M(s_{\vert V})=0$ whenever $V\subset U$ is affine (because $f_M(s_{\vert V})=f_M(s)_{\vert V}$). –  Laurent Moret-Bailly Oct 25 '11 at 6:56

I've surely misunderstood your question or made a simple mistake, but it seems to me that if $X$ is allowed to vary then $Z({\rm QCoh}(X))$ is a sheaf in the Zariski topology---call it $\underline{Z}_X$---and there is a morphism of sheaves $\mathcal{O}_X \rightarrow \underline{Z}_X$. It's an isomorphism in general because it's an isomorphism when $X$ is affine.

Edit: Martin's comment explains the simple mistake I made above, but I think it might be possible to salvage the argument. I'm a little nervous about this though...

Let $A_X$ be the smallest additive subcategory of the category of $\mathcal{O}_X$-modules that contains ${\rm QCoh}(X)$ and is closed under arbitrary products and kernels. Any endomorphism of the identity functor of ${\rm QCoh}(X)$ extends uniquely to an endomorphism of the identity functor of $A_X$ (since product and kernel are functors). Therefore $Z(A_X) = Z({\rm QCoh}(X))$.

Since $f^{-1}$ commutes with arbitrary products and kernels, the categories $A_X$ for varying $X$ form a fibered category (actually a stack) over the Zariski site of $X$.

If $f : U \rightarrow X$ and $G \in A_U$ then $f_\ast G \in A_X$ (see the proof that the pushforward of a quasi-coherent sheaf under a quasi-compact, quasi-separated morphism is quasi-coherent in EGA(1971)I.6.7.1, taking into account that $f_\ast$ commutes with arbitrary inverse limits). If $U$ is an open embedding, this implies that $f^\ast : A_X \rightarrow A_U$ is essentially surjective (since $f^\ast f_\ast G = G$). Therefore any endomorphism of the identity functor of $A_X$ can be restricted to an endomorphism of the identity functor of $A_U$. Therefore as $X$ varies, $Z(A_X) = Z({\rm QCoh}(X))$ forms a sheaf in the Zariski topology.

Edit 2: I have to argue that $f_\ast$ carries $A_U$ into $A_X$. I'm mimicking the argument from EGA here, but I can't see Martin's objection.

We can assume that $X$ is affine (since $A$ is a stack). If $f$ is affine, this is because $f_\ast$ commutes with arbitrary inverse limits and $f_\ast$ carries ${\rm QCoh}(U)$ into ${\rm QCoh}(X)$.

Let $U_i$ be a cover of $U$ by open affines, and let $U_{ij}$ be the pairwise intersections. Let $f_i$ and $f_{ij}$ be the restrictions of $f$. Let $G_i = {f_i}_\ast G_i \big|_{U_i}$ and let $G_{ij} = {f_{ij}}_\ast G \big|_{U_{ij}}$. Then

$f_\ast G = \ker( \prod_i G_i \rightarrow \prod_{ij} G_{ij} )$

and the $G_i$ are in $A_X$ because $U_i \rightarrow X$ are affine. If the $G_{ij}$ are also in $A_X$ then so is $f_\ast G$. This will be the case if $f$ is separated, since then the $U_{ij}$ will be affine. But this implies the general case because the $U_{ij}$ will be quasi-affine, hence separated, over $X$, so $G_{ij} = {f_{ij}}_\ast G \big|_{U_{ij}}$ will be in $A_X$ by the case mentioned above.

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No, the restriction functor does not have to induce a homomorphism between the center. This is only the case if the inclusion is quasi-compact (and this has been dealt before). However, look at mathoverflow.net/questions/38009/… for a context where this actually does make sense. –  Martin Brandenburg Oct 14 '10 at 20:19
Nice construction. I've tried to check the details. It is clear that $Z(A_X) \to Z(Qcoh(X))$ is injective and actually how to define the inverse map. But is it true that this will be well-defined? The endomorphism of a product will be the product of the endomorphisms, but how do we show that these endomorphisms are natural? I think we can repair it when we also make $A_X$ closed under subobjects, but then other problems arise ... What do you think? –  Martin Brandenburg Oct 18 '10 at 9:44
Also there are problems proving that the direct image $A_U \to A_X$ is well-defined. I can reduce it to the case that $U$ is affine. We cannot copy the EGA proof. –  Martin Brandenburg Oct 18 '10 at 16:31
I don't see the problem with the EGA argument, so I'm copying it in above. Can you point out my mistake? Checking naturality seems like a serious problem. How would having subobjects inside $A_X$ help? –  Jonathan Wise Oct 19 '10 at 7:19
The case $X$ affine is clear to me, I should have remarked that. The problem is that in order to show that $A$ is a stack, I actually need the claim about direct images. You seem to use another argument. Which one? –  Martin Brandenburg Oct 19 '10 at 16:01

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