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I have been studying Chapter 8 of Neron models by Bosch et al. The first part deals with the relative Picard functor. A lot of work is done to make it representable. My question would be why this work is done exactly, i.e.:

Why is it useful for the Picard functor to be representable?

I do not have a great overview of this subject, hence I cannot find a use myself.

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  • $\begingroup$ In general, representability is roughly speaking related to the existence of a moduli space for the deformation functor you are considering. $\endgroup$ – Francesco Polizzi Nov 4 '16 at 15:43
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    $\begingroup$ Why were you studying that Chapter 8 if you hadn't been told about a reason why such things are useful? Anyway, look at the chapter (with many historical comments) on Picard schemes by Kleiman in the book "FGA Explained", and contemplate the meaning of Jacobians of smooth proper curves which may not have rational points and how one might relativize that for curves over a base (esp. when some fibers are not smooth). Mazur's work on torsion in elliptic curves over $\mathbf{Q}$ is a glorious application of the latter. $\endgroup$ – nfdc23 Nov 4 '16 at 16:14
  • $\begingroup$ @ndfc23: Thanks, I'll definitely look into FGA Explained. $\endgroup$ – Stapler Nov 4 '16 at 21:12
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    $\begingroup$ It is useful for pretty much any functor to be representable. $\endgroup$ – Sándor Kovács Nov 5 '16 at 4:44
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Let me give one cute example: the Theorem of the Cube (cf. e.g. Mumford's "Abelian varieties") can be proved quite easily if we have Picard schemes at our disposal. In contrast, the proof in Mumford's book is quite complicated. One can even say that the proof below is more enlightening than the one in Mumford, though that one is of course elementary and more general. As far as I understand, the proof of representability of ${\rm Pic}_X$ for $X$ a complete variety does not need the theorem of the cube, so at least the argument below is not circular.

Theorem. Let $(X, x_0)$ and $(Y, y_0)$ be complete pointed varieties (over a fixed algebraically closed field $k$), $(Z, z_0)$ a connected $k$-pointed scheme, $L$ a line bundle on $X\times Y\times Z$ whose restrictions to $\{x_0\}\times Y\times Z$, $X\times \{y_0\}\times Z$, and $X\times Y\times \{z_0\}$ are trivial. Then $L$ is the trivial bundle.

When reading this odd statement for the first time, it is easy to think of the roles of $X$, $Y$, and $Z$ as interchangeable, and that the different assumptions put on $Z$ are probably an artifact of the proof. It is more natural however to think of $L$ as a family of line bundles on $X\times Y$ parametrized by $Z$. So the cube is to be thought of as a `family of squares'...

So if ${\rm Pic}_X$, ${\rm Pic}_Y$, and ${\rm Pic}_{X\times Y}$ are representable (which they are e.g. if $X$ and $Y$ are projective, or maybe even more generally), then $L \in {\rm Pic}_{X\times Y}(Z)$ gives a map $\phi_L:Z\to \underline{\rm Pic}_{X\times Y}$.

The projections $p_X:X\times Y\to X$, $p_Y:X\times Y\to Y$ and the inclusions $i_X:X\cong X\times \{y_0\}\to X\times Y$, $i_Y:Y\cong \{x_0\}\times Y\to X\times Y$ give maps $$ p^*:\underline{\rm Pic}_{X} \times \underline{\rm Pic}_{Y} \to \underline{\rm Pic}_{X\times Y} \quad \text{and}\quad i^*:\underline{\rm Pic}_{X\times Y} \to \underline{\rm Pic}_{X} \times \underline{\rm Pic}_{Y} $$ and one has $i^*\circ p^* = {\rm id}$. If $K_{X, Y}$ denotes the kernel of $p^*$, we get a decomposition $$ \underline{\rm Pic}_{X\times Y} \cong \underline{\rm Pic}_{X} \times \underline{\rm Pic}_{Y} \times K_{X, Y}. $$

Note that our assumptions on $L$ mean exactly that $\phi_L(z_0)=0$ and that $\phi_L$ factors through $K_{X, Y}$!

So far we didn't use the representability, so here it goes: looking at tangent spaces at $0$, by standard deformation theory and the Kunneth formula we see that $$T_0 \underline{\rm Pic}_{X\times Y} \cong H^1(X\times Y, \mathcal{O}_{X\times Y}) \cong H^1(X, \mathcal{O}_X)\times H^1(Y, \mathcal{O}_Y) \cong T_0 \underline{\rm Pic}_{X} \times T_0 \underline{\rm Pic}_{Y}. $$ But by the above product decomposition of $\underline{\rm Pic}_{X\times Y}$, we deduce that $T_0 K_{X, Y} = 0$. Since $K_{X, Y}$ is the kernel of a map between group schemes locally of finite type, it is representable and locally of finite type; since $T_0 K_{X, Y} =0$, $K_{X, Y}$ is etale, and hence discrete (because $k$ was algebraically closed). Since $Z$ is connected, the map $\phi_L : Z\to K_{X, Y}$ is constant, and hence $L$ is trivial.

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