Quasi-coherence of the annihilator ideal sheaf of the sheaf associated to an $A$-module $M$

Question

$\def\Ann{\mathrm{Ann}} \def\Spec{\operatorname{Spec}}$

I am trying to find an example which shows that the annihilator ideal sheaf, denoted by $\mathrm{Ann}(\mathcal F)$, of a quasi-coherent sheaf $\mathcal F$ on a locally-noetherian scheme $X$, is not necessarily quasi-coherent. (I have showed that coherence of $\mathcal F$ implies coherence of $\mathrm{Ann}($$\mathcal F)$.)

The annihilator ideal sheaf is defined by

$\Ann(\mathcal F)(U) = \{f\in O_X(U)|$ $f$ kills $\mathcal F|_U\} =$

$= \{f\in O_X(U)\mid \forall \text{ open } V\subset U$ $f|_V$ kills the $O_X(V)$-module $\mathcal F(V)\}$, for an open $U\subset X$.

For an example, I took the noetherian scheme $X=\Spec(\mathbb{Z})$, the abelian group (or equivalently the $\mathbb{Z}$-module) $M$ to be the subgroup $G\leq\mathbb{Q/Z}$ consisting of all elements whose order is a power of a fixed prime $p$, say $p=2$ (this is an example in Atiyyah&Macdonald, p.74), and looked at the quasi-coherent sheaf $\mathcal F=\widetilde M$, i.e., the sheaf associated to $M$ on $\Spec(\mathbb{Z})$ (for the sheaf associated to a module definition, look in Hartshorne p.110).

Now take $U\subset X$ to be the open set $D(2)=\{q\in X|2\notin q\}$.

It is clear that for every prime ideal $(2)\neq q\in X$ we have $M_q=0$ ($M_q$ is the localization $(\mathbb{Z}-q)^{-1}M$), and thus, $\Ann(\mathcal F)(U) = O_X(U)$. In particular, it follows that $\Ann(\mathcal F)(U)$ is not trivial, since, for example, the section $q\mapsto \frac{1}{1}\in A_q$ is not the zero section in $O_X(U)$.

Why I did do this? because my guess is that if $\Ann(\widetilde M) $ is quasi-coherent, then it will be isomorphic to $\widetilde {\Ann(M)}$, where $\Ann(M)$ is the annihilator of $M$ as a $\mathbb{Z}$-module, but it is easily seen that $\Ann(M)=0$, so [if my guess is correct] we conclude that $\Ann(\widetilde M) $ is not quasi-coherent, as desired.

Do my guess is correct? if it doesn't so, do my example is correct even though?

Answer 1

Whether or not your guess is true, your example is correct. Let $\mathfrak{p}=(2)\in \text{Spec}\mathbb{Z}$. Take $f\in \mathbb{Z}$ such that $f\notin (2)$ (or equivalently, $\mathfrak{p}\in D(f)$), then $M_f=M$, where $M_f$ is the localization of $M$ with respect to the set $\{1,f,f^2,\dots\}$. So $\widetilde{M}(D(f))=M$ whenever $\mathfrak{p}\in D(f)$ and $\widetilde{M}(D(f))=0$ otherwise. It is easy to see that (using the uniqueness of sheaf on a base) $\widetilde{M}$ is a skyscraper sheaf on $\text{Spec}\mathbb{Z}$ concentrates at $\mathfrak{p}$, i.e. $\widetilde{M}(V)$ is $M$ if $\mathfrak{p}\in V$ and $0$ otherwise.

Now consider $Ann(\widetilde{M})$. You have shown that for $U=\text{Spec}\mathbb{Z}-\mathfrak{p}$, $Ann(\widetilde{M})(U)=\mathcal{O}_X(U)$. In fact

$$ Ann(\widetilde{M})_{|U}=\mathcal{O}_U $$ Now take $V$ to be an open neighborhood of $\mathfrak{p}$, $Ann(\widetilde{M})(V)=0$. Hence the stalk of $Ann(\widetilde{M})$ at $\mathfrak{p}$ is $0$. Assume that $Ann(\widetilde{M})$ is quasi-coherent, then it is of finite type. By Lemma 17.9.5 in this Stacks Project entry, there is an open neighborhood $W$ of $\mathfrak{p}$ such that

$$Ann(\widetilde{M})_{|W}=0$$

From this fact, we see that $Ann(\widetilde{M})(W-\mathfrak{p})=0$. But the equation from earlier gives us $Ann(\widetilde{M})(W-\mathfrak{p})=\mathcal{O}_X(W-\mathfrak{p})\neq 0$, a contradiction. We conclude that $Ann(\widetilde{M})$ is not quasi-coherent.