Lemma 39.9.5. An abelian variety is an abelian group scheme, i.e., the group law is commutative.

**Proof.**
Let $k$ be a field. Let $A$ be an abelian variety over $k$. By Lemma 39.9.3 we may replace $k$ by its algebraic closure. Consider the morphism

This is a morphism over $A$ via the first projection on either side. Let $e \in A(k)$ be the unit. Then we see that $h|_{e \times A}$ is constant with value $(e, e)$. By More on Morphisms, Lemma 37.44.3 there exists an open neighbourhood $U \subset A$ of $e$ such that $h|_{U \times A}$ factors through some $Z \subset U \times A$ finite over $U$. This means that for $x \in U(k)$ the morphism $A \to A$, $y \mapsto xyx^{-1}y^{-1}$ takes finitely many values. Of course this means it is constant with value $e$. Thus $(x, y) \mapsto xyx^{-1}y^{-1}$ is constant with value $e$ on $U \times A$ which implies that the group law on $A$ is abelian. $\square$

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