Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a scheme and $x \in X$. Consider the functor

$\text{Qcoh}(X) \to \mathcal{O}_{X,x} \text{-Mod} , M \mapsto M_x.$

Does it have a right-inverse? I.e. is there, for every $\mathcal{O}_{X,x}$-module $N$ a (functorial) quasi-coherent $\mathcal{O}_X$-module $M$ such that there is a natural isomorphism $M_x \cong N$?

If $X$ is quasi-separated, the answer is yes. First observe that, if $X$ is affine, the direct image with respect to $\text{Spec}(\mathcal{O}_{X,x}) \to X$ works. Now if $X$ is quasi-separated, use the affine case to extend $N$ to a quasi-coherent module on an open affine neighborhood $U$ of $x$, and then take the direct image with respect to $U \to X$. This works since $U \to X$ is a quasi-compact, quasi-separated morphism.

In the general case, note that direct images don't work, but this does not disprove the existence of the functor I'm looking for. The question is motivated by this one, which is still unsolved.

If $\mathfrak{m}_x \subseteq \text{rad}(\text{Ann}(N))$, then the direct image with respect to $\text{Spec}(\mathcal{O}_{X,x}/\text{Ann}(N)) \to X$ (this is then an affine morphism!) works.

EDIT: A stronger, but more natural question is the following: Let $U \subseteq X$ an open subset. Does then $res : \text{Qcoh}(X) \to \text{Qcoh}(U)$ have a right inverse? As I said, this is clear if $U \to X$ is a quasi-compact morphism. In general, a transfinite recursion on the length of an affine cover shows that it is enough to consider the case that $X$ is affine, but $U \subseteq X$ arbitrary. Then everything is ok when $U$ is quasi-compact. In general, you can write $U$ as a directed union of quasi-compact subsets of $X$. Does this help somehow?

EDIT 2: There is a right adjoint $\text{Qcoh}(U) \to \text{Qcoh}(X)$ to the restriction functor due to abstract reasons. If $X$ is affine, it maps $N$ to the module associated to $\Gamma(U,N)$, where the latter is considered as a $\Gamma(X,\mathcal{O}_X)$-module. However, this fails to be an extension of $N$, i.e. the counit $\widetilde{\Gamma(U,N)}|_U \to N$ is no isomorphism in general (I have an explicit counterexample). Thus, the desired right-inverse (if it exists) won't be a right adjoint.

EDIT 3: Here is a class of examples where extension works. Assume $X$ is affine and $U \subseteq X$ open can be written as $\coprod_{i \in I} U_i$ with $U_i$ affine. If $M \in \text{Qcoh}(U)$ and $M_i := \Gamma(U_i,M)$, then $\widetilde{\Gamma(U,M)} = \widetilde{\prod_{i \in I} M_i} \in \text{Qcoh}(X)$ is not an extension of $M$, but $\widetilde{\bigoplus_{i \in I} M_i}$ works.

EDIT 4: Let $U,X,M$ as in edit 3. Then an application of Zorn's lemma shows that $M$ can be extended to a maximal open subset $U \subseteq V \subseteq X$. Then for every open affine $W \subseteq X$, either $V \cap W = W$ or $V \cap W$ is not quasi-compact. In particular, $V$ is ("very") dense. For example, $\mathbb{A}^{\infty} - \{0\} \subseteq \mathbb{A}^{\infty}$ is such a dense subset. But I don't know if here extension works. I've already looked at several examples, but have not found out anything ..

EDIT 5: If $dim(X)=0$, then extension works.

Please let me know if you have any ideas!

share|improve this question
Is the answer known if we forget about functoriality? In other words, is every $\mathcal{O}_{X,x}$-module the stalk of some quasicoherent $\mathcal{O}_{X}$-module? –  Laurent Moret-Bailly Sep 7 '10 at 16:23
@Laurent: Perhaps this simplifies the question, but also I don't know how to answer it. –  Martin Brandenburg Sep 7 '10 at 18:02
If $\mathcal{O}_{X,x}$ is a PID and $N$ is finitely generated, then there is an extension $M$ (Reduce to the case $N=\mathcal{O}_{X,x}$ or $N=\mathcal{O}_{X,x} / \mathfrak{m}_x^n$ and use the last remark). –  Martin Brandenburg Sep 8 '10 at 8:33
@Martin: The question in your first edit is exactly the question answered by Grothendieck-style monadic descent. –  Harry Gindi Oct 25 '10 at 1:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.