Surely one could formulate the following question more generally, but as I am primarily concerned with abelian schemes, I will choose this setting:

Let $A$ be an abelian scheme over a locally noetherian base $S$. Each fibre over a point $s$ of $S$ is an abelian variety over the residue field $k(s)$ of $s$.

Now is there a philosophy about which properties of objects on $A$ can be checked fibrewise?

More specifically, I have for example two concrete problems which should illustrate my question:

(1) Given a morphism of coherent sheaves $F\rightarrow G$ on $A$, I can base change this morphism to get $F_s\rightarrow G_s$ a morphism on the fibre $A_s$ over each point $s$ of $S$. If all these base changed morphisms are isomorphisms, can one conclude that the original morphism was an iso too?

(2) Given a coherent module $F$ on $A$, I want to know if it is of the form $e_{*}\mathcal O_S$, where $e$ is the zero section. Can I check this fibrewise?

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    $\begingroup$ For (2), doesn't it even fail for the ab scheme of relative dimension zero? If $A=S$ then aren't you asking "if I have a coherent sheaf on $S$ and I want to check it's $\mathcal{O}_S$, can I check this on points?" and in general of course this is far too much to expect. $\endgroup$ – Kevin Buzzard Jan 2 '12 at 12:09
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    $\begingroup$ 1) and 2) don't have something to do specifically with abelian schemes. Both are almost always wrong. First think about affine schemes ... $\endgroup$ – Martin Brandenburg Jan 2 '12 at 12:10
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    $\begingroup$ Ok, this already helps; I'll think about what additional data I have in my applications. $\endgroup$ – Veen Jan 2 '12 at 13:18
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    $\begingroup$ Why don't you just specify that your coherent sheaves are $S$-flat? $\endgroup$ – Jason Starr Jan 2 '12 at 13:26
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    $\begingroup$ What concerns (2): actually I have the following situation: I know that for every fibre $F_s$ is of the form $e_{*}M_s$ for $M_s$ a module on $Spec(ks))$ (depending on $s$) and I want to conclude that $F$ itself is of the form $e_* M$ for some $\mathcal O_S$-module $M$. $\endgroup$ – Veen Jan 2 '12 at 13:39


I've just realized that I did not read the question carefully, so an edit is in order.

(a) $\mathscr G$ is flat over $S$

So, let's see the real question: Let $\mathscr F\to \mathscr G$ be a morphism of coherent sheaves on $A$ with kernel $\mathscr K$ and cokernel $\mathscr C$. In other words one has an exact sequence $$ 0\to \mathscr K \to \mathscr F \to \mathscr G \to \mathscr C \to 0. $$ Base changing to $k(s)$ gives an exact sequence $$ \mathscr F_s \to \mathscr G_s \to \mathscr C_s \to 0. $$ If $\mathscr F_s \to \mathscr G_s$ is an isomorphism, then $\mathscr C_s=0$ and if this holds for all $s\in S$, then $\mathscr C=0$. (Nakayama's lemma implies that if $\mathscr C_s=0$, then $\mathscr C$ is zero along $f^{-1}(s)$, hence $\mathrm{supp} \mathscr C\subset A\setminus f^{-1}(s)$. If this holds for all $s\in S$, then $\mathrm{supp}\mathscr C=\emptyset$, so $\mathscr C=0$).

So now we have a short exact sequence $$ 0\to \mathscr K \to \mathscr F \to \mathscr G \to 0. $$ Base changing to $k(s)$ gives the exact sequence $$ \mathscr K_s \to \mathscr F_s \to \mathscr G_s \to 0, $$ and if $\mathscr G$ is flat over $S$, then this last exact sequence is also exact on the left. In that case, again, if $\mathscr F_s \to \mathscr G_s$ is an isomorphism, then $\mathscr K_s=0$ and if this holds for all $s\in S$, then $\mathscr K=0$ (same way as above).

(b) $\mathscr G$ is not flat over $S$

Here is a simple example to show that $\mathscr G$ being flat (or something else perhaps) is a necessary condition. Let $S$ be arbitrary, $E$ an abelian variety and $A=E\times S$. Let $p\in E$, $t\in S$, and $Z=\{p\}\times S$. Further let $\mathfrak m\subseteq \mathscr O_Z$ be the maximal ideal corresponding to the point $(p,t)\in Z$. Now let $\mathscr F=\mathscr O_Z/\mathfrak m^2$, $\mathscr G=\mathscr O_Z/\mathfrak m$, and $\mathscr F \to \mathscr G$ the natural projection. Then $\mathscr F_s \to \mathscr G_s$ is an isomorphism for every $s\in S$: Indeed if $s\neq t$, then $\mathscr F_s = \mathscr G_s=0$ and for $s=t$, they are both isomorphic to $\mathscr O_Z/\mathfrak m$ and the restriction of the above morphism is an isomorphism, but the original $\mathscr F \to \mathscr G$ is not injective.

(c) other stuff

Originally I did not read the question carefully and thought you just want to obtain information about $\mathscr F$ and $\mathscr G$ from how $\mathscr F_s$ and $\mathscr G_s$ behave. In other words, if there weren't a morphism $\mathscr F \to \mathscr G$ given, then the outlook is much more bleak. Actually, in part (2) it is not clear whether you have such a morphism or not, so this might still be relevant for that case.

If you do not have a morphism $\mathscr F \to \mathscr G$ given, you would always have to account for twisting one of the sheaves by a line bundle pulled back from $S$: For any $s\in S$, $\mathscr F_s\simeq (\mathscr F\otimes f^*\mathscr N)_s$ where $f:A\to S$ is the structure map and $\mathscr N$ is a line bundle on $S$.

There is something along these lines that is actually true:
Assume that $S$ is integral of finite type over an algebraically closed field, $f$ is flat and projective and the fibers are integral. Let $\mathscr L$ and $\mathscr M$ be two line bundles on $A$ such that $\mathscr L_s\simeq \mathscr M_s$ for all $s\in S$. Then there exists a line bundle $\mathscr N$ on $S$ such that $\mathscr L\simeq \mathscr M\otimes \mathscr N$.

To prove this, apply cohomology and base change to the sheaf $f_*(\mathscr L\otimes \mathscr M^{-1})$. I suppose you can try to generalize the method of proving this to a little bit more general situation, but you probably need more information to get much further.


Let $Z=e(S)$. Then $f|_Z:Z\to S$ is an isomorphism and in particular $\mathscr O_Z$ is flat over $S$. So, if you have a morphism $\mathscr F\to \mathscr O_Z$, then you can apply (1). If you don't have such a morphism, then there is the problem I mentioned above.

  • $\begingroup$ Aren't all maps of line bundles which are isomorphisms at each point, isomorphisms? If the first bundle is trivial and the second is not, then the global section of the trivial bundle must go to a global section of the other, which must be 0 somewhere. So look at the fiber there. Otherwise, tensor to make the first bundle trivial. Is there a hidden condition in this argument that I'm forgetting about? Is it just wrong? $\endgroup$ – Will Sawin Jan 2 '12 at 22:46
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    $\begingroup$ Will, you are right. I've just realized that I have misread the original question so I edited my answer. I think it resolves the issue you're mentioning as well. $\endgroup$ – Sándor Kovács Jan 3 '12 at 2:22
  • $\begingroup$ Thank you for your nice answer and your effort editing it, Sándor. $\endgroup$ – Veen Jan 3 '12 at 8:52

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