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 $S$.
Best Answer
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.