The Stacks project

Lemma 39.9.10. Let $k$ be a field of characteristic $p > 0$. Let $A$ be an abelian variety of dimension $g$ over $k$. The fibre of $[p] : A \to A$ over $0$ has at most $p^ g$ distinct points.

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

\[ [p]^{-1}(U) \xrightarrow {\pi _1} \mathop{\mathrm{Spec}}(\mathcal{O}(U)[x_1, \ldots , x_ g]/(x_1^ p - t_1, \ldots , x_ g^ p - t_ g)) \xrightarrow {\pi _2} U \]

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. $\square$

Comments (2)

Comment #7865 by James on

Here it could be noted that .

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