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Let $X = \operatorname{Spec} k[x]_{(x)}$ which consists of two elements, the generic point $\zeta$ corresponding to the zero ideal and the closed point $(x)$. Define an $\mathcal{O}_X$-module $\mathcal{F}$ by setting $\mathcal{F}(X) = \{0\}$ and $\mathcal{F}(\zeta) = k(x).$ Now $\mathcal{F}$ is not a quasi-coherent sheaf because if $\mathcal{F}|_{\operatorname{Spec} k[x]_{(x)}} = \mathcal{F}$ is isomorphic to $\widetilde{M}$ for some $A$-module $M$, $\mathcal{F}(X) = 0$ implies that $\widetilde{M}(X) = M = 0$. But now $\mathcal{F}(\zeta)$ cannot be isomorphic to $\widetilde{M}(\zeta)$ because one is non-zero while the other is zero. Thus $\mathcal{F} \notin \operatorname{QCoh}(X)$.

Are there any other examples of $\mathcal{O}_X$-modules that are not quasi-coherent sheaves?

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  • $\begingroup$ By $\mathcal{O}_X$ do you mean that $X=$Spec $k[x]_{(x)}$? $\endgroup$
    – edo arad
    Aug 14, 2013 at 8:15
  • $\begingroup$ @edoarad Yes, $\mathcal{O}_X$ is the structure sheaf of $X = \text{Spec} k[x]_{(x)}$. $\endgroup$
    – user38268
    Aug 14, 2013 at 8:52

3 Answers 3

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Let $A$ be a discrete valuation ring (for example your $\operatorname{Spec} k[x]_{(x)}$), $\;X=\operatorname{Spec} (A)=\{\zeta, f\}$ the corresponding affine scheme ( $f$ the closed point , $\zeta $ the generic point ) and $U=\{\zeta\}$ the unique non empty and non full open subset of $X$.

An $ \mathcal{O}_X$-module $\mathcal{F}$ consists of an $A$-module $M (=\mathcal F(X))$, a $K=Frac(A)$-module $N(=\mathcal F(U))$ and an $A$-linear map $M\to N$ corresponding to the sheaf restriction.
[Note the amusing and unusual fact that every presheaf on $X$ is automatically a sheaf since $X$ has no non-trivial covering!]
These data automatically give rise to a canonical morphism of $K$-vector spaces $$F:M\otimes_AK\to N $$ Characterization of quasi-coherence
The sheaf $\mathcal F$ is quasi-coherent if and only if $F$ is bijective.

And now you can boast that you can describe all quasi-coherent sheaves on $X$ !

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  • $\begingroup$ Dear Georges, thanks for your answer. I see now that my example above would've given rise to a whole family of examples! Are there examples of $\mathcal{O}_X$-modules that are not quasi-coherent sheaves when $X$ is $\operatorname{Spec} A$, $A$ not a DVR? $\endgroup$
    – user38268
    Aug 14, 2013 at 8:50
  • $\begingroup$ Of course there are lots of examples. $\endgroup$ Aug 14, 2013 at 9:06
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    $\begingroup$ Dear @Benja, I have answered your request in a separate post, for clarity. $\endgroup$ Aug 14, 2013 at 9:21
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    $\begingroup$ I suppose @Zhen alludes to the value of the presheaf on the empty set. Those interested in such empty considerations should add the condition that the presheaf is zero on the empty set or make some suitable modification in order to appease their conscience. $\endgroup$ Aug 14, 2013 at 11:13
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    $\begingroup$ I agree with Zhen here. Presheaves don't have to map $\emptyset$ to the terminal object. This is only the case for sheaves. The definition of a presheaf in Hartshorne's book is wrong and shows once again that one should consult other books as for the foundations of algebraic geometry. $\endgroup$ Aug 15, 2013 at 7:28
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Since @Benja asks, here is an example of a non-quasi-coherent sheaf $\mathcal I$ on the spectrum of a ring which is not a discrete valuation ring, namely $X=Spec (k[T])=\mathbb A^1_k$ ($k$ a field):

Consider the origin $O\in X=\mathbb A^1_k$ correponding to the maximal ideal $(T)\subset k[T]$ and define the ideal subsheaf $\mathcal I(U)\subset \mathcal O_X(U)$ by:

$\begin{cases} \mathcal I(U)= \mathcal O_X(U)\;\text {if}\; O\notin U \\ \mathcal I(U)= 0\;\text {if} \; O\in U \end{cases} $

The sheaf $\mathcal I$ is not quasi-coherent because $\mathcal I\neq 0$ although $\mathcal I(X)=0$.

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    $\begingroup$ Note that this is $j_! \mathcal{O}_U$, where $j : U := X \setminus \{0\} \hookrightarrow X$. $\endgroup$ Aug 14, 2013 at 10:51
  • $\begingroup$ For reference, this follows from the lemma: stacks.math.columbia.edu/tag/01IA $\endgroup$
    – 54321user
    Mar 6, 2017 at 4:35
  • $\begingroup$ It seems $\mathcal I$ is just presheaf so it needs sheafification, say $\mathcal I^{"}$ .Then is $\mathcal I^{"}(X) = 0$ still true ? $\endgroup$
    – hew
    Jan 22, 2021 at 8:33
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    $\begingroup$ @hew: No, $\mathcal I$ is already a sheaf and doesn't need sheafification. Why do you (wrongly) believe otherwise? $\endgroup$ Jan 22, 2021 at 8:46
  • $\begingroup$ @GeorgesElencwajg , I was wrong. Usually, $j_!$ contains the concept of sheafification, but in this case, I see now it becomes the sheaf as it is. Thank you. $\endgroup$
    – hew
    Jan 22, 2021 at 9:05
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Let $R$ be a discrete valuation ring and $X=\mathrm{Spec}(R)$. Hence, as a set, we have $X=\{\eta,x\}$. The topology is such that $x$ is closed, but $\eta$ is not (in other words, it is the Sierpinski space). The structure sheaf is given by $\mathcal{O}(\emptyset)=0$, $\mathcal{O}(\{\eta\})=\mathcal{O}_{\eta} = K$, the field of fractions of $R$, and $\mathcal{O}(X)=R$. An $\mathcal{O}$-module corresponds to an $R$-module $A$ (global sections) and an $K$-module $B$ (sections on $\{\eta\}$) equipped with a homomorphism of $R$-modules $A \to B|_R$ (restriction), or equivalently a homomorphism of $K$-modules $\vartheta : A \otimes_R K \to B$. It is quasi-coherent iff $\vartheta$ is an isomorphism. This gives lots of examples of $\mathcal{O}$-modules which are not quasi-coherent. For example, $\vartheta$ could be zero, etc.

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