Examples of $\mathcal{O}_X$-modules that are not quasi-coherent sheaves

Question

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?

Answer 1

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.

Answer 2

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 $X$ !

Answer 3

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 $\mathcal I(X)=0$.