# The Picard scheme

*Posted on April 24, 2016 by Dima*

## 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’’)