# Picard functor is not linear

Let $S$ be a locally noetherian scheme, $Y$ a locally noetherien $S$-scheme and $X$ an abelian scheme over $S$. It is known that the map between groups $Hom(Y,X) \to Hom(Pic(X/S),Pic(Y/S)), f \mapsto f^*$ is quadratic, i.e. we have

$(f+g+h)^* - (f + g)^* - (f + h)^* - (h + h)^* + f^* + g^* + h^* = 0$.

However, $f \mapsto f^\*$ is not linear. Is there an easy example for $(f+g)^* \neq f^* + g^*$?

This is related to the order of the functor $X \mapsto Pic(X/S)$. The above inequality would include a nontrivial line bundle on $X \times_S X$, which is trivial on the both closed subschemes $X \times 0, 0 \times X$. I'm also interested in an easy example for this phenomenon.

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The universal example would be $Y=X$, $f=g=id$. The map is linear on $Pic^0$, but quadratic on $Pic$. As for your second paragraph, the Poincare bundle on the square of an abelian variety is an easy example . – t3suji Jun 30 '10 at 14:21
Martin, complex-analytically (with $S = {\rm{Spec}}(\mathbb{C})$), the component group of the Picard scheme is naturally a subgroup of ${\rm{H}}^2(X, \mathbb{Z}) = \wedge^2({\rm{H}}^1(X,\mathbb{Z}))$, so you can see the non-linear quadratic nature of the component group quite "visibly" there. – Boyarsky Jun 30 '10 at 14:25

Let $E$ be an elliptic curve; take $f = g = \mathrm{id}_E$. Then $f+g$ is multiplication by $2$, and has degree $4$; hence if $L$ is an invertible sheaf on $E$, the sheaf $(f+g)^*L$ has degree $4 \deg L$, while $f^*L \otimes g^*L = L^{\otimes 2}$ has degree $2\deg L$.