Let \(X\) be an integral projective scheme over an algebraically closed field \(k\), and assume that \(H^1(X, \mco_X) = 0\). Let \(T\) be a connected scheme of finite type over \(k\).
If \(\mcl\) is an invertible sheaf on \(X \times T\), show that the invertible sheaves \(\mcl_t\) on \(X = X \times \ts{t}\) are isomorphic for all closed points \(t \in T\).
Show that \(\Pic(X \times T) = \Pic X \times \Pic T\). Do not assume that \(T\) is reduced.
Hint: apply (12.11) with \(i = 0, 1\) for suitable invertible sheaves on \(X \times T\).
Cf. (IV, Ex. 4.10) and (V, Ex. 1.6) for examples where \(\Pic(X \times T) \neq \Pic X \times \Pic T\).
Solution
Let \[ p:X\times T\longrightarrow X, \qquad q:X\times T\longrightarrow T \] be the two projections. Since \(X\) is projective over \(k\), the morphism \(q\) is projective. It is also flat, being the base change of \(X\to\Spec k\).
Because \(X\) is integral and projective over the algebraically closed field \(k\), \[ H^0(X,\mco_X)=k. \] By hypothesis, \[ H^1(X,\mco_X)=0. \]
1. Local rigidity lemma. Let \(\mcf\) be an invertible sheaf on\(X\times T\), and let \(t\in T\) be a closed point such that\[ \mcf_t\cong\mco_X. \]Then there is an open neighborhood \(U\) of \(t\) and an invertible sheaf\(\mcn_U\) on \(U\) such that\[ \mcf|_{X\times U}\cong q_U^*\mcn_U, \]where \(q_U:X\times U\to U\) is the projection.
The sheaf \(\mcf\) is flat over \(T\), because it is locally isomorphic to \(\mco_{X\times T}\), which is flat over \(T\).
Consider the degree-one base-change map at \(t\): \[ \beta_t^1: (R^1q_*\mcf)\tensor k(t) \longrightarrow H^1(X,\mcf_t). \] Since \(t\) is closed, \(k(t)=k\), and \[ H^1(X,\mcf_t) \cong H^1(X,\mco_X) = 0. \] Thus \(\beta_t^1\) is surjective. By the full cohomology-and-base-change theorem, Hartshorne III.12.11, after shrinking around \(t\), the maps \(\beta_s^1\) are isomorphisms. In particular \[ (R^1q_*\mcf)\tensor k(t)=0. \] Nakayama’s lemma gives \[ (R^1q_*\mcf)_t=0. \] Since \(R^1q_*\mcf\) is coherent, shrink once more so that \[ R^1q_*\mcf=0 \] on a neighborhood of \(t\).
The degree-one criterion in III.12.11 now implies that the degree-zero base-change map \[ \beta_t^0: (q_*\mcf)\tensor k(t) \longrightarrow H^0(X,\mcf_t) \] is surjective. Applying III.12.11 in degree zero, after another shrinking the sheaf \[ \mcn_U=(q_U)_*(\mcf|_{X\times U}) \] is locally free and degree-zero base change is an isomorphism. Its rank is \[ h^0(X,\mcf_t) = h^0(X,\mco_X) = 1, \] so \(\mcn_U\) is invertible.
There is a canonical evaluation map \[ \epsilon:q_U^*\mcn_U\longrightarrow\mcf|_{X\times U}. \] On the fibre over \(t\), base change identifies it with \[ H^0(X,\mco_X)\tensor\mco_X \longrightarrow \mco_X, \] which is an isomorphism. Let \(\mcc\) be its cokernel. The support of \(\mcc\) is closed in \(X\times U\), and its image under the projective morphism \(q_U\) is closed in \(U\). This image does not contain \(t\). After deleting it, \(\epsilon\) is surjective. Both sides are invertible sheaves, so a surjection between them is an isomorphism. This proves the lemma.
2. The isomorphism class of \(\mcl_t\) is locally constant as \(t\)ranges over the closed points of \(T\).
Fix a closed point \(t\in T\), and set \[ \mca=\mcl_t\in\Pic X. \] On \(X\times T\), put \[ \mcf=\mcl\tensor p^*(\mca^{-1}). \] Its fibre at \(t\) is trivial: \[ \mcf_t \cong \mcl_t\tensor\mca^{-1} \cong \mco_X. \] By step 1, there is an open neighborhood \(U_t\) of \(t\) on which \(\mcf\) is pulled back from \(U_t\). Therefore for every closed point \(s\in U_t\), \[ \mcf_s\cong\mco_X, \] and hence \[ \mcl_s\cong\mca\cong\mcl_t. \] Thus the fibre isomorphism class is locally constant on closed points.
3. All the closed-point fibres \(\mcl_t\) are mutually isomorphic.This proves part (a).
For every isomorphism class \(\alpha\in\Pic X\) that occurs among the closed fibres, let \(V_\alpha\subseteq T\) be the union of the neighborhoods \(U_t\) from step 2 over closed points \(t\) satisfying \[ [\mcl_t]=\alpha. \] Each \(V_\alpha\) is open.
If \(V_\alpha\cap V_\gamma\ne\varnothing\), then this nonempty open subset of the finite-type \(k\)-scheme \(T\) contains a closed point \(s\). The defining property of the two neighborhoods gives \[ [\mcl_s]=\alpha=\gamma. \] Hence the distinct \(V_\alpha\) are pairwise disjoint.
Their union contains every closed point of \(T\). Its complement is a closed subset with no closed points. Since a finite-type scheme over a field is Jacobson, that complement is empty. Thus the \(V_\alpha\) form a disjoint open covering of \(T\). Because \(T\) is connected, only one of them is nonempty. Therefore \[ \mcl_t\cong\mcl_s \] for all closed points \(s,t\in T\), proving (a).
4. Suppose an invertible sheaf \(\mcf\) on \(X\times T\) satisfies\[ \mcf_t\cong\mco_X \]for every closed point \(t\in T\). Then there is an invertible sheaf\(\mcn\) on \(T\) such that\[ \mcf\cong q^*\mcn. \]
Set \[ \mcn=q_*\mcf. \] Fix a closed point \(t\in T\). The same III.12.11 argument as in step 1 applies because \[ H^0(X,\mcf_t)=k, \qquad H^1(X,\mcf_t)=0. \] It gives an open neighborhood \(U_t\) on which \(\mcn\) is locally free of rank \(1\), and on which degree-zero base change is an isomorphism.
The union of these \(U_t\) contains every closed point of \(T\). The non-locally-free locus of the coherent sheaf \(\mcn\) is closed; if it were nonempty, it would contain a closed point. Hence \(\mcn\) is invertible on all of \(T\).
Now consider the global evaluation map \[ \epsilon:q^*\mcn\longrightarrow\mcf. \] For every closed \(t\), degree-zero base change identifies its restriction to \(X\times\{t\}\) with \[ H^0(X,\mco_X)\tensor\mco_X \longrightarrow \mco_X, \] so it is an isomorphism on every closed fibre.
Let \(\mcc=\operatorname{coker}(\epsilon)\). If \(\operatorname{Supp}\mcc\) were nonempty, then, being a closed subset of the finite-type \(k\)-scheme \(X\times T\), it would contain a closed point \((x,t)\). Its image \(t\) is then a closed point of \(T\), but the restriction of \(\epsilon\) to that fibre is surjective, a contradiction. Thus \(\mcc=0\), so \(\epsilon\) is surjective. Since its source and target are invertible sheaves, it is an isomorphism: \[ \mcf\cong q^*\mcn. \] This argument used III.12.11 and nowhere assumed that \(T\) is reduced.
5. Every invertible sheaf on \(X\times T\) is a tensor product ofpullbacks from the two factors.
Let \(\mcl\) be invertible on \(X\times T\), and choose a closed point \(t_0\in T\). Put \[ \mca=\mcl_{t_0}\in\Pic X \] and \[ \mcf=\mcl\tensor p^*(\mca^{-1}). \] By part (a), for every closed \(t\), \[ \mcl_t\cong\mcl_{t_0}=\mca, \] hence \[ \mcf_t\cong\mco_X. \] Step 4 supplies an invertible sheaf \(\mcn\) on \(T\) with \[ \mcf\cong q^*\mcn. \] Therefore \[ \boxed{\mcl\cong p^*\mca\tensor q^*\mcn}. \] This proves surjectivity of the natural homomorphism \[ \Phi:\Pic X\times\Pic T\longrightarrow\Pic(X\times T), \qquad ([\mca],[\mcn]) \longmapsto [p^*\mca\tensor q^*\mcn]. \]
6. The homomorphism \(\Phi\) is injective.
Suppose \[ p^*\mca\tensor q^*\mcn\cong\mco_{X\times T}. \] Restrict to the fibre over any closed point \(t\in T\). The restriction of \(q^*\mcn\) is a trivial line bundle, so \[ \mca\cong\mco_X. \] Thus \[ q^*\mcn\cong\mco_{X\times T}. \]
Choose a closed point \(x\in X\). Since \(k\) is algebraically closed, \(x\) is \(k\)-rational. It defines a section \[ s_x:T\longrightarrow X\times T, \qquad t\longmapsto(x,t) \] of \(q\). Pulling the last isomorphism back by \(s_x\) gives \[ \mcn\cong\mco_T. \] Hence the kernel of \(\Phi\) is trivial.
7. Therefore\[ \boxed{\Pic(X\times T)\cong\Pic X\times\Pic T}. \]
Step 5 proves surjectivity of \(\Phi\), and step 6 proves injectivity. This proves part (b), including the case of nonreduced connected \(T\).
8. Q.E.D.
Steps <1>1–<1>3 prove part (a), and steps <1>4–<1>7 prove part (b).