Lemma 39.9.8. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $[d] : A \to A$ be the multiplication by $d$. Then $[d]$ is finite locally free of degree $d^{2\dim (A)}$.

Proof. By Lemma 39.9.2 (and More on Morphisms, Lemma 37.46.1) 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 $C \subset X$ 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 33.43.14 for example. (You can also use Varieties, Lemma 33.44.9.) 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 33.44.11 we see that

$\deg _{\mathcal{N}^{\otimes d^2}}(A) = \deg ([d]) \deg _\mathcal {N}(A)$

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

$n \longmapsto \chi (A, \mathcal{N}^{\otimes nd^2}),\quad \text{respectively}\quad n \longmapsto \chi (A, \mathcal{N}^{\otimes n})$

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

Comment #5572 by anon on

There is a typo at the end of the first paragraph: $\mathcal{N}^{\otimes n^2}$ should be $\mathcal{N}^{\otimes d^2}$.

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