Nonzero multiplication on an abelian variety is finite and faithfully flat

Choose a projective embedding line bundle $H$ on $A$, and put $N=H\otimes [-1{]}^{\ast }H$. The inverse is an automorphism, so $[-1{]}^{\ast }H$ is also an embedding line bundle. The product of the two embeddings followed by Segre restricts along the closed diagonal $A\to A\times A$ to a closed immersion with line bundle $N$, by [F2]. Thus $N$ is very ample and symmetric, since inversion interchanges its factors. By [F1], $[n{]}^{\ast }N\cong N^{\otimes n^2}$, which is ample by [F2]. The map $[n]$ is proper: its graph is closed because $A$ is separated, and the projection $A\times A\to A$ is proper as a base change of $A\to \mathrm{Spec}k$.

Each geometric fibre of $[n]$ is proper. On any reduced irreducible component $C$ of such a fibre, $N^{\otimes n^2}{\mathrm{\mid }}_C$ is trivial, since it is pulled back from the point to which that fibre maps; it is also ample by [F2]. A high power therefore defines a closed immersion of $C$ into projective space. But $C$ is a proper integral variety over the algebraically closed geometric ground field, so every section of this trivial power is a scalar by [F3]. Its projective map is constant, and a constant closed immersion forces $C$ to be a point. Thus each geometric fibre is zero-dimensional and has finitely many points, since it is Noetherian with finitely many irreducible components. By [F3], $[n]$ is quasi-finite and hence finite. This argument uses $n^2>0$ as an integer, regardless of its image in $k$.

The finite morphism is surjective. Its image is closed; factor through its reduced image $B$, an integral closed subvariety of $A$, since $A$ is integral and the ring kernels defining the image are prime. On affine charts the resulting finite maps from nonempty source charts are injective integral ring extensions; thus [F3], applied to the finite affine preimages and their common function-field dimensions, gives $\dim B=\dim A=g$. A proper closed subset of an irreducible finite-dimensional variety has smaller dimension: any irreducible chain in that subset extends by appending the entire variety. Hence $B=A$. This also applies after every field extension, by the same finite-fibre proof; in particular over an algebraic closure $K$ every $y\in A(K)$ has a closed rational point in its nonempty finite fibre by [F4].

Over $K$, [F4] supplies a dense open $U$ of the target on which $[n]$ is flat. Surjectivity makes its inverse image nonempty; choose a rational point $x_0$ there. For any other rational point $x$, translation by $x-x_0$ on the source and by $[n](x-x_0)$ on the target intertwine $[n]$, because it is a homomorphism. These automorphisms carry the flat local ring map at $x_0$ to the one at $x$, so $[n]$ is flat at every closed point of the source. Equivalently, over a closed target point $y$, all localizations of the finite module algebra $([n{]}_{\ast }{\mathcal{O}}_A{)}_y$ at its maximal ideals are flat over ${\mathcal{O}}_{A,y}$; these maximal ideals are precisely the finitely many closed source points over $y$. This implies the whole module is flat: a kernel of $I{\otimes }_{R}B\to B$ localizes to zero at every maximal ideal of $B$, for each finitely generated ideal $I$ of $R={\mathcal{O}}_{A,y}$, and hence is zero. The module is finite, thus free by [F4]. Its locally free locus is open. The complementary closed subset has no closed points, so is empty by [F4]. Therefore $[n{]}_K$ is finite locally free. On each affine chart the extension to $K$ is faithfully flat over $k$, and [F4] descends flatness of the finite algebra; finite flat algebras over the Noetherian target are locally free. Together with surjectivity in step 3.1 this proves faithful flatness over $k$.

Let $r$ be the rank of $E=[n{]}_{\ast }{\mathcal{O}}_A$, constant since $A$ is connected. Finite morphisms are affine, so their inverse images of affine covers and intersections are affine. The Čech complexes give $\chi (A,[n{]}^{\ast }{N}^t)=\chi (A,E\otimes N^t)$: on each chart this is the elementary projection formula $B{\otimes }_{R}P$ for a line module $P$. Write $P(t)=\chi (A,N^t)$. By [F5], $P$ has degree $g$ and nonzero leading coefficient $c$. Also the coefficient of $t^g$ in $\chi (A,E\otimes N^t)$ is $rc$: choose $a$ with $E\otimes N^a$ globally generated, and choose $r$ global sections forming a basis at the generic point. They give an injection ${\mathcal{O}}_A^{\oplus r}\to E\otimes N^a$, since $A$ is integral and the kernel of a generically injective map from a free sheaf is zero. Its cokernel $Q$ has support in a proper closed subset, hence dimension less than $g$; [F5] gives a polynomial of degree less than $g$ for $Q$. Twisting and additivity give $\chi (E\otimes N^t)=rP(t-a)+\chi (Q\otimes N^{t-a})$, with leading coefficient $rc$. Since step 1.1 gives $[n{]}^{\ast }{N}^t=N^{n^{2}t}$, the left side of the Čech identity is $P(n^{2}t)$, with leading coefficient $c\mathrm{\mid }n{\mathrm{\mid }}^{2g}$. Cancelling $c\mathrm{\ne }0$ proves $r=\mathrm{\mid }n{\mathrm{\mid }}^{2g}$. For $g=0$, the same argument has $Q=0$ and compares constant polynomials.

Base change of a finite locally free map preserves its rank. The identity fibre is therefore a finite locally free $k$-scheme of rank $\mathrm{\mid }n{\mathrm{\mid }}^{2g}$; the group operations restrict to this fibre because $[n]$ is a homomorphism, giving $A[n]$. For $a:T\to A$, the pullback $T^{\mathrm{\prime }}=T{\times }_{a,A,[n]}A\to T$ is finite faithfully flat, and its projection to $A$ supplies an $n$th root of $a{\mathrm{\mid }}_{T^{\mathrm{\prime }}}$. Over an algebraically closed field, step 3.1 gives a rational point in each fibre, proving rational-point surjectivity. All arguments apply in characteristic dividing $n$; none use the differential of $[n]$ or assert that the map or its kernel is étale.