The Stacks project

Proof. This follows from Lemma 39.8.7 and More on Morphisms, Lemma 37.50.1. $\square$

Proof. Omitted. Note that this is why we insisted on $A$ being geometrically integral; without that condition this lemma (and many others below) would be wrong. $\square$

Proof. If $k$ is perfect then this follows from Lemma 39.8.2 (characteristic zero) and Lemma 39.8.4 (positive characteristic). We can reduce the general case to this case by descent for smoothness (Descent, Lemma 35.23.27) and going to the perfect closure using Lemma 39.9.3. $\square$

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. $\square$

Proof. Apply the theorem of the cube (More on Morphisms, Theorem 37.33.8) to the difference

This works because the restriction of $\mathcal{M}$ to $A \times A \times e = A \times A$ is equal to

where $n_{i_1, \ldots , i_ t} : A \times _ k A \to A$ is the morphism $(x_1, x_2) \mapsto \sum x_{i_ j}$. Similarly for $A \times e \times A$ and $e \times A \times A$. $\square$

Proof. Consider the morphism $A \to A \times _ k A \times _ k A$, $x \mapsto (x, x, -x)$ where $-x = [-1](x)$. Pulling back the relation of Lemma 39.9.6 we obtain

which proves the result for $n = 2$. By induction assume the result holds for $1, 2, \ldots , n$. Then consider the morphism $A \to A \times _ k A \times _ k A$, $x \mapsto (x, x, [n - 1]x)$. Pulling back the relation of Lemma 39.9.6 we obtain

and the result follows by elementary arithmetic. $\square$

Proof. By Lemma 39.9.2 (and More on Morphisms, Lemma 37.50.1) we see that $A$ has an ample invertible module $\mathcal{L}$. Since $[-1] : A \to A$ is an automorphism, we see that $[-1]^*\mathcal{L}$ is an ample invertible $\mathcal{O}_ X$-module as well. Thus $\mathcal{N} = \mathcal{L} \otimes [-1]^*\mathcal{L}$ is ample, see Properties, Lemma 28.26.5. Since $\mathcal{N} \cong [-1]^*\mathcal{N}$ we see that $[d]^*\mathcal{N} \cong \mathcal{N}^{\otimes d^2}$ by Lemma 39.9.7.

To get a contradiction, let $C \subset X$ be a proper curve contained in a fibre of $[d]$. Then $\mathcal{N}^{\otimes d^2}|_ C \cong \mathcal{O}_ C$ is an ample invertible $\mathcal{O}_ C$-module of degree $0$ which contradicts Varieties, Lemma 33.44.14 for example. (You can also use Varieties, Lemma 33.45.9.) Thus every fibre of $[d]$ has dimension $0$ and hence $[d]$ is finite for example by Cohomology of Schemes, Lemma 30.21.1. Moreover, since $A$ is smooth over $k$ by Lemma 39.9.4 we see that $[d] : A \to A$ is flat by Algebra, Lemma 10.128.1 (we also use that schemes smooth over fields are regular and that regular rings are Cohen-Macaulay, see Varieties, Lemma 33.25.3 and Algebra, Lemma 10.106.3). Thus $[d]$ is finite flat hence finite locally free by Morphisms, Lemma 29.48.2.

Finally, we come to the formula for the degree. By Varieties, Lemma 33.45.11 we see that

Since the degree of $A$ with respect to $\mathcal{N}^{\otimes d^2}$, respectively $\mathcal{N}$ is the coefficient of $n^{\dim (A)}$ in the polynomial

we see that $\deg ([d]) = d^{2 \dim (A)}$. $\square$

Proof. Observe that $[d](x + y) = [d](x) + [d](y)$. Since translation by a point is an automorphism of $A$, we see that the set of points where $[d] : A \to A$ is étale is either empty or equal to $A$ (some details omitted). Thus it suffices to check whether $[d]$ is étale at the unit $e \in A(k)$. Since we know that $[d]$ is finite locally free (Lemma 39.9.8) to see that it is étale at $e$ is equivalent to proving that $\text{d}[d] : T_{A/k, e} \to T_{A/k, e}$ is injective. See Varieties, Lemma 33.16.8 and Morphisms, Lemma 29.36.16. By Lemma 39.6.4 we see that $\text{d}[d]$ is given by multiplication by $d$ on $T_{A/k, e}$. $\square$

Proof. To prove this, we may and do replace $k$ by the algebraic closure. By Lemma 39.6.4 the derivative of $[p]$ is multiplication by $p$ as a map $T_{A/k, e} \to T_{A/k, e}$ and hence is zero (compare with proof of Lemma 39.9.9). Since $[p]$ commutes with translation we conclude that the derivative of $[p]$ is everywhere zero, i.e., that the induced map $[p]^*\Omega _{A/k} \to \Omega _{A/k}$ is zero. Looking at generic points, we find that the corresponding map $[p]^* : k(A) \to k(A)$ of function fields induces the zero map on $\Omega _{k(A)/k}$. Let $t_1, \ldots , t_ g$ be a p-basis of $k(A)$ over $k$ (More on Algebra, Definition 15.46.1 and Lemma 15.46.2). Then $[p]^*(t_ i)$ has a $p$th root by Algebra, Lemma 10.158.2. We conclude that $k(A)[x_1, \ldots , x_ g]/(x_1^ p - t_1, \ldots , x_ g^ p - t_ g)$ is a subextension of $[p]^* : k(A) \to k(A)$. Thus we can find an affine open $U \subset A$ such that $t_ i \in \mathcal{O}_ A(U)$ and $x_ i \in \mathcal{O}_ A([p]^{-1}(U))$. We obtain a factorization

of $[p]$ over $U$. After shrinking $U$ we may assume that $\pi _1$ is finite locally free (for example by generic flatness – actually it is already finite locally free in our case). By Lemma 39.9.8 we see that $[p]$ has degree $p^{2g}$. Since $\pi _2$ has degree $p^ g$ we see that $\pi _1$ has degree $p^ g$ as well. The morphism $\pi _2$ is a universal homeomorphism hence the fibres are singletons. We conclude that the (set theoretic) fibres of $[p]^{-1}(U) \to U$ are the fibres of $\pi _1$. Hence they have at most $p^ g$ elements. Since $[p]$ is a homomorphism of group schemes over $k$, the fibre of $[p] : A(k) \to A(k)$ has the same cardinality for every $a \in A(k)$ and the proof is complete. $\square$

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

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

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 $0 \leq f \leq g$. $\square$