If \(X\) is an elliptic curve, show that there is an exact sequence \[ 0 \to p_1^* \Pic X \oplus p_2^* \Pic X \to \Pic(X \times X) \to R \to 0, \] where \(R=\operatorname{End}(X, P_0)\). In particular, we see that \(\Pic(X \times X)\) is bigger than the sum of the Picard groups of the factors. Cf. (III, Ex. 12.6), (V, Ex. 1.6).
Solution
Write \(O=P_0\), and let \[ p_1,p_2:X\times X\longrightarrow X \] be the projections. We regard \(R=\Endo(X,O)\) as an additive group.
1. The map\[ \Pic X\oplus\Pic X \longrightarrow \Pic(X\times X), \qquad (\mca,\mcb)\longmapsto p_1^*\mca\tensor p_2^*\mcb \]is injective.
Suppose \[ p_1^*\mca\tensor p_2^*\mcb\cong\OO_{X\times X}. \] Restricting to \(X\times\{O\}\) gives \(\mca\cong\OO_X\), because the second factor restricts to a constant one-dimensional vector space tensored with \(\OO_X\). Restricting to \(\{O\}\times X\) then gives \(\mcb\cong\OO_X\). Thus the kernel is zero.
2. Every line bundle \(\mcl\) on \(X\times X\) canonically determines an endomorphism \(q(\mcl)\in\Endo(X,O)\).
Let \[ i_2:X\longrightarrow X\times X, \qquad y\longmapsto(O,y), \] and put \[ \mcm=i_2^*\mcl, \qquad \mcl^0=\mcl\tensor p_2^*\mcm^{-1}. \] Then \(i_2^*\mcl^0\cong\OO_X\).
The degree of the restriction of a line bundle to the fibres of \(p_1:X\times X\to X\) is constant on the connected base \(X\). Indeed, \(\mcl^0\) is flat over \(X\), and Theorem III.9.9 says that a flat projective family has constant Hilbert polynomial. For an ample invertible sheaf \(\mch\) on the second factor, Riemann–Roch on the genus-one fibre gives \[ \chi(\mcl^0_x\tensor\mch^{\tensor n}) = \deg\mcl^0_x+n\deg\mch, \] so constancy of the Hilbert polynomial forces \(\deg\mcl^0_x\) to be constant. Since the fibre over \(O\) is trivial for \(\mcl^0\), every fibre restriction has degree zero. The relative Picard construction therefore gives a morphism \[ \phi_{\mcl}:X\longrightarrow\Pic^0(X), \qquad x\longmapsto \left[\mcl^0|_{\{x\}\times X}\right]. \] It sends \(O\) to the identity. Under \[ \alpha_X:X\overset\sim\longrightarrow\Pic^0(X), \qquad P\longmapsto\OO_X(P-O), \] define \[ q(\mcl)=\alpha_X^{-1}\circ\phi_{\mcl}:X\longrightarrow X. \] This morphism sends \(O\) to \(O\), hence is an endomorphism of the elliptic curve.
3. The assignment\[ q:\Pic(X\times X)\longrightarrow R \]is a homomorphism, and every bundle of the form\[ p_1^*\mca\tensor p_2^*\mcb \]lies in its kernel.
For line bundles \(\mcl,\mcn\), normalization commutes with tensor product, and on every fibre of \(p_1\), \[ (\mcl\tensor\mcn)^0_x \cong \mcl^0_x\tensor\mcn^0_x. \] Tensor product in \(\Pic^0(X)\) corresponds under \(\alpha_X\) to addition on \(X\). Hence \[ q(\mcl\tensor\mcn)=q(\mcl)+q(\mcn). \]
If \(\mcl=p_1^*\mca\tensor p_2^*\mcb\), then normalization by the fibre over \(O\) removes the \(p_2^*\mcb\) factor, up to a trivial constant factor. The restriction of \(p_1^*\mca\) to every fibre \(\{x\}\times X\) is trivial. Therefore \(q(\mcl)=0\).
4. Conversely,\[ \ker q = p_1^*\Pic X\oplus p_2^*\Pic X. \]
2. Then\[ \left[\mcl^0|_{\{x\}\times X}\right]=0 \qquad \text{in }\Pic^0(X) \]for every \(x\in X\).Thus \(\mcl^0\) represents the zero element of the relative Picard group for \(p_1:X\times X\to X\).
By the defining quotient in the relative Picard functor, a line bundle representing the zero relative class is pulled back from the base. Hence there is \(\mcn\in\Pic X\) such that \[ \mcl^0\cong p_1^*\mcn. \] Undoing the normalization gives \[ \mcl \cong p_1^*\mcn\tensor p_2^*\mcm. \] Step 3 gives the reverse inclusion, proving the equality.
5. The homomorphism \(q\) is surjective.
Let \(f\in\Endo(X,O)\). In the proof of Exercise IV.4.7(d), the normalized relative Picard bundle \[ \mcm_f = \OO_{X\times X}\bigl(\Gamma_f-X\times\{O\}\bigr) \tensor p_1^*f^*\OO_X(O)^{-1} \] was constructed. Its restriction to the fibre \(\{x\}\times X\) is \[ \OO_X(f(x)-O). \] Therefore the morphism to \(\Pic^0(X)\) associated to \(\mcm_f\) is precisely \[ x\longmapsto\OO_X(f(x)-O)=\alpha_X(f(x)). \] By the definition of \(q\) in step 2, \[ q(\mcm_f)=f. \] Thus every element of \(R\) is in the image.
6. Q.E.D.
Step 1 gives injectivity on the left, steps <1>3–<1>4 identify the kernel of \(q\) with the two pullback Picard groups, and step 5 proves surjectivity. Hence \[ 0 \longrightarrow p_1^*\Pic X\oplus p_2^*\Pic X \longrightarrow \Pic(X\times X) \xrightarrow{q} \Endo(X,O) \longrightarrow 0 \] is exact. Since \(R\) contains the nonzero identity endomorphism, the quotient of \(\Pic(X\times X)\) by the two pullback Picard groups is nonzero. Thus the pullback subgroup is proper, which is the stated final consequence.