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Let $A$ be an abelian variety over a field $k$ and let $K_A$ be its canonical divisor. Then I'm almost certain that $K_A$ is trivial, but I can't seem to prove it, nor find a counter example, nor find any reference on abelian varieties that even mentions canonical divisors.

Does anyone know a reference saying (or an idea for a short proof) that the canonical divisor on an abelian variety is trivial?

I'd prefer not to assume anything about $k$, but a result with $\operatorname{char}(k) = 0$ would be better than nothing.

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up vote 9 down vote accepted

The tangent bundle of any group variety is trivial: take a basis of the tangent space at any one point and translate it around using the group law.

Therefore the cotangent bundle of an abelian variety, being the dual of a trivial vector bundle, is trivial. This implies -- but is in general much stronger than! -- that the top exterior power of the cotangent bundle is trivial.

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