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!!

Can you help me, please?

  • 2
    $\begingroup$ 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!! $\endgroup$ Feb 1, 2015 at 17:40

1 Answer 1


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 $V(p) \simeq \mathrm{Spec} \ \mathbb Z/p\mathbb Z$.

Edit: I should mention, btw, that the definition you give for support is different than the definition given in Hartshorne (which is, I think, the accepted reference for these things). The definition in Hartshorne is that the support is simply the set of points at which the stalk is nonzero. What you've stated here will be the closure of that set. It's often the case, as in this example, that Hartshorne's support is closed and therefore coincides with what you have here, but in general that needn't be the case.


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