# The Picard scheme

Posted on April 24, 2016 by Dima

tags: moduli, Abelian varieties

## Functors $$Pic$$ and $$Div$$

We will denote base change with a subscript: $$X_T = X \times T$$.

If $$X$$ is a scheme, the Picard group of $$X$$ is defined to be the group of isomorphism classes of invertible sheaves on $$X$$. The relative Picard functor of an $$S$$-scheme $$X$$ is defined as ${\operatorname{Pic}}_{X/S}(T) := {\operatorname{Pic}}(X_T) / {\operatorname{Pic}}(T)$ where the embedding $${\operatorname{Pic}}(T) \hookrightarrow {\operatorname{Pic}}(X_T)$$ is given by the pullback along the structure maps $$X_T \to T$$.

Caveat: all representability results work with a sheafification of this functor in some topology, Zariski, étale, or fppf. Unless $$X \to S$$ is proper and has a section, they need not be isomorphic.

An effective divisor is a closed subscheme such that its ideal is invertible. If $$f: X \to S$$ is a morphism of schemes then a relative effective divisor on $$X$$ is an effective divisor $$D$$ such that $$D$$ is flat over $$S$$. For a morphism $$X \to S$$ define the functor of relative divisors ${\operatorname{Div}}_{X/S}(T) := {\{\ \textrm{ relative effective divisors on } X_T/T \ \}}$

We are interested in representability of this functor, so to this end we prove a little lemma.

Lemma. Let $$X \to S$$ be a flat morphism. Let $$D$$ be a closed subscheme of $$X$$ flat over $$S$$. Then $$D$$ is relative effective divisor in a neighbourhood of $$x \in X$$ if and only if $$D_s$$ is cut out in a neighbourhood of $$x$$ in $$X_s$$, where $$s$$ is the image of $$x$$, by a single non-zero element of $${\operatorname{\mathcal{O}}}_{X_s,x}$$ (which amounts to being an effective divisor in $$X_s$$ but we won’t prove it).

Proof. Left to right. Multiplication by the element $$f$$ that cuts out $$D$$ induces the short exact sequence of $${\operatorname{\mathcal{O}}}_{S,s}$$-modules $0 \to {\operatorname{\mathcal{O}}}_{X,x} \to {\operatorname{\mathcal{O}}}_{X,x} \to {\operatorname{\mathcal{O}}}_{D,x} \to 0$ Tensoring it with $$k(s)$$ we get $\begin{array}{c} \ldots \to {\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{X,x}, k(s)) \to {\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{X,x}, k(s)) \to {\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{D,x}, k(s)) \to \\ \to {\operatorname{\mathcal{O}}}_{X,x} \otimes k(s) \to {\operatorname{\mathcal{O}}}_{X,x} \otimes k(s) \to {\operatorname{\mathcal{O}}}_{D,x} \otimes k(s) \to 0\\ \end{array}$ Since $$D$$ is flat in a neighbourhood of $$x$$, the third term $${\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{D,x}, k(s))$$ vanishes. Therefore, the element $$f \otimes k(s)$$ cuts out $$D_s$$ in $$X_s$$, and $$D_s$$ is an effective divisor in $$X_s$$.

Right to left: suppose $$D_s$$ is locally cut out by non-zero divisor of $${\operatorname{\mathcal{O}}}_{X_s,x}$$ then we have to show $$D$$ is locally cut out by a single element of $${\operatorname{\mathcal{O}}}_{X,x}$$.

Consider the exact secquence $0 \to I_{D,x} \to {\operatorname{\mathcal{O}}}_{X,x} \to {\operatorname{\mathcal{O}}}_{D,x} \to 0$ then the long exact sequence associated to tensoring with $$k(s)$$ is ${\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{X,x}, k(s)) \to {\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{D,x}, k(s)) \to I_{D,x} \otimes k(s) \to {\operatorname{\mathcal{O}}}_{X,x} \otimes k(s) \to {\operatorname{\mathcal{O}}}_{D,x} \otimes k(s) \to 0$ as $$I_{D,x} = f{\operatorname{\mathcal{O}}}_{X,x}$$ and $$f$$ is regular in the fibre $$X_s$$, $$I_{D,x} \otimes k(s) \to {\operatorname{\mathcal{O}}}_{X,x} \otimes k(s)$$ is multiplication by $$f \otimes k(s)$$, so in fact is an inclusion. $${\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{X,x}, k(s))$$ vanishies sinche $$X$$ is flat over $$S$$, and therefore, finally, $${\operatorname{Tor}}_1({\operatorname{\mathcal{O}}}_{D,x}, k(s))=0$$, and $$D$$ is flat over $$S$$.

Since, as it follows from the premise, the multplication by $$f \otimes k(s)$$ induces an isomorphism $${\operatorname{\mathcal{O}}}_{X,x} \otimes k(s) \to I_{D,x} \otimes k(s)$$, it follows by Nakayama’s lemma that the kernel of the multiplication by $$f$$ is trivial, and so $$f$$ is not a zero divisor.

Corollary. $${\operatorname{Div}}$$ is representable by an open subscheme of $${\operatorname{Hilb}}$$.

Proof. Let $$W \subset X \times {\operatorname{Hilb}}_{X/S}$$ be the universal scheme. Then the property of being an effective divisor on $$X \times {\operatorname{Hilb}}$$ in a neighourhood of a given point is an open property because of the previous lemma, let $$U \subset W$$ be the set of such points. Then since projection to $${\operatorname{Hilb}}_{X/S}$$ is proper, the image of $$U$$ in $${\operatorname{Hilb}}_{X/S}$$ is open. Let us show that it represents $${\operatorname{Div}}_{X/S}$$. Note that for any $$y \in O$$, $$W_y$$ is an effective divisor in $$X_y$$ by the Lemma above.

Given relative effective divisor $$D \subset X_T$$ there exists a map $$\iota_D: T \to {\operatorname{Hilb}}_{X/S}$$ such that $$D = W \times_{{\operatorname{Hilb}},\iota_D} T$$. At every point $$t \in {\operatorname{Im}}(\iota_D: T \to {\operatorname{Hilb}})$$, the closed subscheme $$W_t$$ is a divisor in $$X_t$$, and by the Lemma above $$W_T$$ is a relative effective divisor.

## representability of $${\operatorname{Pic}}(X/S)$$, Mumford’s method

Suffices to represent $${\operatorname{Pic}}^\tau$$, the component of $${\operatorname{Pic}}$$ that contains numerically trivial divisors, because components corresponding to different numerical classes are isomorphic. In fact, it will be more convenient to fix a very ample divisor $$\xi$$ and represent the compononet $${\operatorname{Pic}}^\xi$$ of $${\operatorname{Pic}}$$ consisting of divisors numerically equivalent to $$\xi$$. We pick $$\xi$$ to be 0-regular (regularity was described in the previous post).

There is a little semi-tautological argument that it is enough to show representability of $${\operatorname{Pic}}^\xi$$ (given that $${\operatorname{Div}}^\xi$$ is representable).

We consider the morphism of fuctors $$\Phi: {\operatorname{Div}}\to {\operatorname{Pic}}$$ and then restrict it to $${\operatorname{Pic}}^\xi$$. The goal is to construct a section $$s: {\operatorname{Pic}}^\xi \to {\operatorname{Div}}^\xi$$.

Lemma. Assume $$s$$ exists, then $${\operatorname{Pic}}$$ is representable.

Proof. The morphism $$s \circ \Phi$$ is an endomorphism of $${\operatorname{Div}}^\xi$$, which is representable, say by an $$S$$-scheme $$D$$. Then $$s \circ \Phi({\operatorname{id}}_D)$$ is an endomorphism $$f$$ of the scheme $$D$$. Consider the fibre product $$D \times_{\Delta, D \times D, {\operatorname{id}}\times f} D$$, where $$\Delta: D \to D \times_S D$$ is the diagonal map. Call this fibre product $$P$$.

We claim that $$P$$ represents $${\operatorname{Pic}}$$.

Indeed, $${\operatorname{Hom}}(T, P)$$ is isomorphic, by construction, to the set of pairs of morphsims $$\alpha, \beta: T \to D$$ such that $$\Delta(\alpha) = {\operatorname{id}}\times f (\beta)$$, i.e. $${\operatorname{Hom}}(T, P)$$ is the image of $${\operatorname{Hom}}(T, D)$$ under $$s \circ \Phi$$. This means that $$P$$ represents $${\operatorname{Pic}}^\xi$$. $$\square$$

The section is constructed as follows (Mumford assumes $$S=k$$, a field, not sure how this is restrictive; on the other hand he also assumes $$X$$ is a surface, this seems to be crucial later).

Suppose we are given an invertible sheaf $${\operatorname{\mathcal{L}}}$$ no $$X \times T$$. Denote $${\operatorname{\mathcal{M}}}_x$$ its restriction to $${x} \times T$$, which can be considered as a sheaf on $$T$$, and let $${\operatorname{\mathcal{E}}}$$ be the sheaf of global sections $$(p_T)_* {\operatorname{\mathcal{L}}}$$.

We pick a finite number of points $$x_1, \ldots, x_N \in X$$, and consider the natural morphism $h: {\operatorname{\mathcal{E}}}\to \bigoplus_{i=1}^N {\operatorname{\mathcal{M}}}_{x_i}, s \mapsto (s(x_1), \ldots, s(x_N))$ Assuming the rank of $${\operatorname{\mathcal{E}}}$$ is $$r$$ (and this is the same for invertible sheaves of the same numerical class, if this class is sufficiently ample), we can wedge this $$r-1$$ times to get a homomorphism of invertible sheaves $(\wedge h)^*: \otimes {\operatorname{\mathcal{M}}}_{x_i} \to {\operatorname{Hom}}(\wedge {\operatorname{\mathcal{E}}}, {\operatorname{\mathcal{O}}}_T)$ This gives a canonical morphism of sheaves ${\operatorname{\mathcal{O}}}_T \to {\operatorname{\mathcal{E}}}\otimes ((\wedge {\operatorname{\mathcal{E}}})^{-1} \otimes (\otimes M_{x_i}))$ and hence a canonical section $\sigma \in H^0(X \times T, {\operatorname{\mathcal{L}}}\otimes (p_T)^* ((\wedge {\operatorname{\mathcal{E}}})^{-1} \otimes (\otimes M_{x_i}))$ Using some magick related to regularity (takes part in the choice of $$\xi$$), one can show that the zero locus of this section does not vanish on any fibre $$X_t$$.

(All this is Lectures 19-20 in Lectures on curves on algebraic surface’’)