Lemma 39.9.7. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ A$-module. Then

\[ [n]^*\mathcal{L} \cong \mathcal{L}^{\otimes n(n + 1)/2} \otimes ([-1]^*\mathcal{L})^{\otimes n(n - 1)/2} \]

where $[n] : A \to A$ sends $x$ to $x + x + \ldots + x$ with $n$ summands and where $[-1] : A \to A$ is the inverse of $A$.

**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

\[ \mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [-1]^*\mathcal{L} \cong [2]^*\mathcal{L} \]

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

\[ [n + 1]^*\mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [n - 1]^*\mathcal{L} \cong [2]^*\mathcal{L} \otimes [n]^*\mathcal{L} \otimes [n]^*\mathcal{L} \]

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

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