5
$\begingroup$

Given $n$ a non-zero integer, it is known that multiplication by $n$ on an abelian variety (defined over any field $k$) is an isogeny. The proof of this fact uses the existence of an ample symmetric divisor on these varieties, which are projective.

Is this statement also true in general for abelian schemes, which may not be projective ?

I know that it is true for elliptic curves (as schemes), as it is proved in Katz and Mazur's book. However, the proof also makes use of the projectivity of such curves and their concrete description in terms of a Weierstraß equation.


For reference, an abelian scheme $X$ over a base scheme $S$ is a smooth proper $S$-group scheme with geometrically connected fibers. A homomorphism $f:X\rightarrow Y$ (as $S$-group schemes) of abelian schemes is an isogeny if it is surjective with a finite kernel. By "finite", we mean that the kernel is an $S$-group scheme which is locally free of finite rank over $S$. When the base is noetherian, this is just a finite flat group scheme over $S$.

$\endgroup$
2
  • $\begingroup$ Just to say that Munford defines abelian variety as a complete group variety, that it is an abelian group is a theorem which follows from that the conjugaison $x\to gxg^{-1}$ induces a regular map from $G$ complete to $Aut(Lie(G))$ affine so it must be constant. $\endgroup$ – reuns Feb 11 '20 at 1:25
  • $\begingroup$ @reuns Yes, that is absolutely true. Abelian schemes are also automatically commutative in virtue of the rigidity theorem (as stated in Mumford's GIT). Fibers of abelian schemes naturally are abelian varieties. $\endgroup$ – Suzet Feb 11 '20 at 1:29
4
$\begingroup$

Without loss of generality we may assume that the base scheme $S$ is Noetherian. Let $X/S$ be an abelian scheme and let $[n]:X \to X$ be the multiplication by $n$ map. Since $X/S$ is smooth it is in particular flat, and a morphism between flat $S$-schemes is flat if and only if its fibers are flat (fibral criterion of flatness). But if $s \in S$ then $[n]:X_s \to X_s$ is just the multiplication by $[n]$ map on an abelian variety over a field, which is flat. Moreover, properness of $X$ implies properness of $[n]$ and once again we can check that $[n]$ is quasi-finite on fibers, and then proper+quasi finite implies finite.

To conclude, $[n]$ is a finite flat morphism and so the kernel is a finite flat group schemes over $S$. Surjectivity can be checked fiberwise as well, so $[n]$ is an isogeny.

$\endgroup$
2
  • $\begingroup$ Thank you very much for these explanations, that's exactly what I needed. For what reason is it enough to treat the case where $S$ is Noetherian ? Is it some sort of general principle ? $\endgroup$ – Suzet Feb 11 '20 at 23:00
  • 2
    $\begingroup$ EDIT : After some digging, I see that reduction to the Noetherian case can be done thanks to some good results found in EGA IV. Thank you again ! $\endgroup$ – Suzet Feb 12 '20 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.