Computation of the support of a sheaf

Shafarevich defines the support of a sheaf $\mathcal{F}$ as the set $X \setminus M$ where $M=\bigcup_{V \subset X} V$ ($V$ open) with $\mathcal{F}(U)=0$ for all nonempty $U \subset V$.

After some pages, he asks the following question: is $\overline{\mathcal{F}}$ a coherent sheaf on $Y$, or even a sheaf of $\mathcal{O}-$modules? (where $Y$ is a reduced closed subscheme of $X$ and $\overline{\mathcal{F}}$ is the sheaf obtained after the devissage process)

He says no, providing a counterexample; that's to say if $X=\operatorname{Spec}(\mathbb{Z})$ and $\mathcal{F}$ is the coherent sheaf corresponding to the module $\mathbb{Z}/p^2\mathbb{Z}$, with $p$ prime.

At this point he writes that the support of this sheaf is the prime ideal generated by $p$ and the corresponding reduced subscheme is $\operatorname{Spec(\mathbb{Z}/p\mathbb{Z})}$. But I don't know how his calculations work!!

• The stalk of this sheaf at a point $x$ corresponding to the prime ideal $<q>$ is the localisation of $\mathbb{Z}/p^2\mathbb{Z}$ with respect to $\mathbb{Z} \setminus P$, where $P$ is a prime ideal not containing $x$. So every stalk is the set $\{\frac{[y]_{p^2}}{z},$ where $z$ is not divisible by $x \}$. Now, I have to proof that the stalk at $x$ is $0$ if and only if $x \ne <p>$, but I don't know how I can do it!! Feb 1, 2015 at 17:40

Let $X = \mathrm{Spec} \ \mathbb Z$. The points are $(0)$ and $(q)$ for primes $q$.
First what are the stalks of your sheaf? We take the SES $$\mathbb Z \overset{p^2}{\to} \mathbb Z \to \mathbb Z/p^2\mathbb Z$$ and tensor with $\mathbb Z_{(0)} = \mathbb Q$ or $\mathbb Z_{(q)}$ and look at what the last module is. For $\mathbb Q$ and $\mathbb Z_{(q)}$ (when $q \neq p$) mult by $p^2$ is an isomorphism (because $p$ becomes a unit in the localization) and localization is exact so the stalk is $0$. For $\mathbb Z_{(p)}$ it's not true that $p^2$ is a unit so the stalk is $\mathbb Z_{(p)}/p^2\mathbb Z_{(p)} \neq 0$.
Now consider the closed set $V(p)$ (which is just the point $(p)$). On the complement $X \setminus V(p)$ all the stalks are zero so the sheaf, restricted to this open set, is zero. That means the complement of the support contains $X \setminus V(p)$. It cannot contain anything more because the stalk at $(p)$ is nonzero, so on any open set containing $(p)$ the sheaf is nonzero. This means the complement of the support is exactly $X \setminus V(p)$, so the support is $V(p)$.
Basic scheme stuff then gives $V(p) \simeq \mathrm{Spec} \ \mathbb Z/p\mathbb Z$.