Let $A$ be an abelian scheme over a base scheme $S$ and $\omega$ a global section of the differential module $\Omega^1\_{A\times\_S A/S}$.


Suppose that $\omega$ is zero when restricted to $A\times S$ and $S\times A$, both times via the zero section and the identity.


Then why can one conclude that $\omega$ itself is already zero?



Differential forms on abelian schemes
It has nothing to do with abelian schemes. Just use the ``product rule'' of differentiation.
This is a natural isomorphism:


$$
p\_1^\*\Omega^1\_{A/S}\oplus p\_2^\*\Omega^1\_{A/S}\to \Omega^1\_{A\times\_S A/S}
$$


The zero section of $A$ affords an inverse to this natural map: if $s\_1 : A\to A\times\_S A$ is the inclusion in the first coordinate and $s\_2$ the analogous inclusion in the second coordinats, then since $p\_is\_i = id$ we see that $s\_1\oplus s\_2 : \Omega^1\_{A\times\_S A}\to p\_1^\*\Omega^1\_{A/S}\oplus p\_2^\*\Omega^1\_{A/S}$ is the inverse of the above isomorphism.


