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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?

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up vote 5 down vote accepted

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.

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