For a line bundle $L$ on $X$, define the <homomorphism associated to a line bundle on an abelian variety>
$$
\phi_L:X(k)\longrightarrow\operatorname{Pic}(X),
\qquad
\phi_L(x)=T_x^*L\otimes L^\vee.
$$
The <Theorem of the square> gives
$$
\phi_L(x+y)=T_{x+y}^*L\otimes L^\vee\simeq\phi_L(x)\otimes\phi_L(y),
$$
so $\phi_L$ is a homomorphism. Pullback distributes over the <tensor product of sheaves>, and therefore
$$
\boxed{\phi_{L\otimes M}(x)=\phi_L(x)\otimes\phi_M(x).}
$$
Iterating the homomorphism law in $x$ gives
$$
\boxed{\phi_{L^{\otimes n}}(x)=\phi_L(x)^{\otimes n}=\phi_L(nx).}
$$

Suppose $L^{\otimes n}\in\operatorname{Pic}^0(X)$. Then $\phi_L(nx)$ is trivial for every $x$. The <multiplication-by-n morphism> on an abelian variety is surjective, so $\phi_L$ is trivial and $L\in\operatorname{Pic}^0(X)$. Thus the <Néron-Severi group>
$$
\operatorname{Pic}(X)/\operatorname{Pic}^0(X)
$$
is torsion-free.

Finally, put $M=\phi_L(x)$. For every $y$,
$$
\phi_M(y)=T_y^*(T_x^*L\otimes L^\vee)\otimes(T_x^*L\otimes L^\vee)^\vee
\simeq T_{x+y}^*L\otimes T_y^*L^\vee\otimes T_x^*L^\vee\otimes L,
$$
which is trivial by the Theorem of the square. Hence
$$
\boxed{\operatorname{im}\phi_L\subseteq\operatorname{Pic}^0(X).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 126 / 4 / ii / Solution

{title2=$[n]_G$}

For a commutative <group scheme> $G$, the multiplication-by-$n$ morphism $[n]_G:G\to G$ adds a point to itself $n$ times. On an <abelian variety>, it is surjective for every nonzero integer $n$.

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 Incoming links: Multiplication-by-n morphism