Definition 39.9.1. Let $k$ be a field. An abelian variety is a group scheme over $k$ which is also a proper, geometrically integral variety over $k$1.

[1] For equivalent definitions see Remark 39.9.12.

Comment #78 by Keenan Kidwell on

The section title should be Picard groups of curves, right?

Comment #7838 by Zhenhua Wu on

There are $2\times 4\times 2=16$ equivalent definitions of abelian varieties. Let {projective, proper}, {geometrically irreducible, irreducible, geometrically connected, connected}, {smooth, geometrically reduced} be three sets of properties, pick one from each of them, and let $A/K$ be a group scheme with the chosen properties. Then it agrees with the definition of abelian variety. I suggest we add this somewhere just for the sake of new students who read too many books and get confused with the different definitions of abelian varieties. Actually I have the following proof in my thesis.

Proof: Clearly projectivity implies properness; and geometrically irreducibility implies the other three which in turn implies connectivity. By tag 056T we know that smoothness of K-schemes implies geometrically reducedness. Thus it suffices to show that "abelian varieties are projective, geometrically irreducible and smooth", and "proper connected and geometrically reduced group schemes over $K$ is an abelian variety".

By tag 0BF9, abelian varieties are projective and smooth. Geometrically irreducibility comes from the definition. Next we show that proper connected and geometrically reduced group schemes over $K$ are abelian varieties, i.e. they are proper and geometrically integral.

It suffices to show geometrically irreducibility. A group scheme over the field $K$ must contain a $K$-rational point, the unit section. In this case, $A$ is connected implies it is geometrically connected by tag 04KV. Hence after base change to any field extension $K′$ of $K$, $A_{K′}$ is connected. Plus, since every connected group scheme over a field is irreducible by tag 0B7Q, $A_{K′}$ is irreducible. So $A$ is geometrically irreducible. The result follows.

Comment #8117 by Zhenhua Wu on

Cool! But have you added this remark on the site? I don't see it in this section.

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