Wonderfully explained in [AVar].

Proposition 39.9.11. Let $A$ be an abelian variety over a field $k$. Then

1. $A$ is projective over $k$,

2. $A$ is a commutative group scheme,

3. the morphism $[n] : A \to A$ is surjective for all $n \geq 1$,

4. if $k$ is algebraically closed, then $A(k)$ is a divisible abelian group,

5. $A[n] = \mathop{\mathrm{Ker}}([n] : A \to A)$ is a finite group scheme of degree $n^{2\dim A}$ over $k$,

6. $A[n]$ is étale over $k$ if and only if $n \in k^*$,

7. if $n \in k^*$ and $k$ is algebraically closed, then $A(k)[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2\dim (A)}$,

8. if $k$ is algebraically closed of characteristic $p > 0$, then there exists an integer $0 \leq f \leq \dim (A)$ such that $A(k)[p^ m] \cong (\mathbf{Z}/p^ m\mathbf{Z})^{\oplus f}$ for all $m \geq 1$.

Proof. Part (1) follows from Lemma 39.9.2. Part (2) follows from Lemma 39.9.5. Part (3) follows from Lemma 39.9.8. If $k$ is algebraically closed then surjective morphisms of varieties over $k$ induce surjective maps on $k$-rational points, hence (4) follows from (3). Part (5) follows from Lemma 39.9.8 and the fact that a base change of a finite locally free morphism of degree $N$ is a finite locally free morphism of degree $N$. Part (6) follows from Lemma 39.9.9. Namely, if $n$ is invertible in $k$, then $[n]$ is étale and hence $A[n]$ is étale over $k$. On the other hand, if $n$ is not invertible in $k$, then $[n]$ is not étale at $e$ and it follows that $A[n]$ is not étale over $k$ at $e$ (use Morphisms, Lemmas 29.36.16 and 29.35.15).

Assume $k$ is algebraically closed. Set $g = \dim (A)$. Proof of (7). Let $\ell$ be a prime number which is invertible in $k$. Then we see that

$A[\ell ](k) = A(k)[\ell ]$

is a finite abelian group, annihilated by $\ell$, of order $\ell ^{2g}$. It follows that it is isomorphic to $(\mathbf{Z}/\ell \mathbf{Z})^{2g}$ by the structure theory for finite abelian groups. Next, we consider the short exact sequence

$0 \to A(k)[\ell ] \to A(k)[\ell ^2] \xrightarrow {\ell } A(k)[\ell ] \to 0$

Arguing similarly as above we conclude that $A(k)[\ell ^2] \cong (\mathbf{Z}/\ell ^2\mathbf{Z})^{2g}$. By induction on the exponent we find that $A(k)[\ell ^ m] \cong (\mathbf{Z}/\ell ^ m\mathbf{Z})^{2g}$. For composite integers $n$ prime to the characteristic of $k$ we take primary parts and we find the correct shape of the $n$-torsion in $A(k)$. The proof of (8) proceeds in exactly the same way, using that Lemma 39.9.10 gives $A(k)[p] \cong (\mathbf{Z}/p\mathbf{Z})^{\oplus f}$ for some $0 \leq f \leq g$. $\square$

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