What is the center of Qcoh(X)? - MathOverflow

What is the center of Qcoh(X)?

2010-10-11T14:25:17Z

The center of a category $C$ is the monoid $Z(C)=\text{End}_{C^C}(1_C)$. 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 a $Ab$-category, this is actually a ring. For example, the center of $\text{Mod}(A)$ is the center of $A$, if $A$ is a (noncommutative) ring.

Now my question is: What is the center of $\text{Qcoh}(X)$, where $X$ is a scheme? Observe that there is a natural map $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X))$; a global section is mapped to the endomorphisms of the quasi-coherent modules which are given by multiplication with this section. Also, there is a natural map $Z(\text{Qcoh}(X)) \to \Gamma(\mathcal{O}_X,X)$, which takes a compatible family of endomorphisms to the image of the global section $1$ in $\mathcal{O}_X$. The composite $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X)) \to \Gamma(\mathcal{O}_X,X)$ 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 $\Gamma(\mathcal{O}_X,X) \to Z(\text{Qcoh}(X))$ 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 of $X$. Since Rosenberg generalized this to arbitrary schemes, it is tempting to look at his proof. But if I understand correctly, Rosenberg uses a structure sheaf on the spectrum of an abelian category which avoids the above problems and uses $Z(\text{Mod}(X))=\Gamma(\mathcal{O}_X,X)$, which is certainly true (use extensions by zero). But I'm not sure because Rosenberg refers to a proof step (a4) which is not there ...

Edit: Angelo has proven it below if $X$ is quasi-separated. Now what happens if $X$ is not quasi-separated?

Answer by Angelo for What is the center of Qcoh(X)?

2010-10-11T15:28:54Z

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.

Answer by Jonathan Wise for What is the center of Qcoh(X)?

2010-10-14T17:48:10Z

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.