Multiplication by an integer on an abelian variety is an etale morphism if and only if the integer is invertible in the base field.

Lemma 39.9.9. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d] : A \to A$ is étale if and only if $d$ is invertible in $k$.

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 29.35.16. By Lemma 39.6.4 we see that $\text{d}[d]$ is given by multiplication by $d$ on $T_{A/k, e}$. $\square$

Comment #3035 by Brian Lawrence on

Suggested slogan: Multiplication by an integer on an abelian variety is an etale morphism if and only if the integer is invertible in the base field.

Comment #5364 by BCnrd on

Should assume $A\ne 0$ (so $T_{A/k,e} \ne 0$) or else "only if" is not true (fails when $A=0$). Maybe also replace "$d$" with "$m$", to avoid awkward-looking notation such as ${\rm{d}} .

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