Let \(Y\) be an integral scheme of finite type over an algebraically closed field \(k\). Let \(\mce\) be a locally free sheaf on \(Y\), and let \(X = \PP(\mce)\); see (II, §7).
Show that \(\Pic X \cong (\Pic Y) \times \ZZ\). This strengthens (II, Ex. 7.9).
(Erratum: the rank-one case)
The displayed formula requires \[ \operatorname{rank}\mce\ge2, \] as in Exercise II.7.9. If \(\operatorname{rank}\mce=1\), then \(\PP(\mce)\cong Y\), so \[ \Pic\PP(\mce)\cong\Pic Y \] rather than \(\Pic Y\times\ZZ\) in general.
The solution below proves the corrected statement for \(\operatorname{rank}\mce\ge2\).
Solution
Let \[ p:X=\PP(\mce)\longrightarrow Y \] be the projection, let \[ r=\operatorname{rank}\mce\ge2, \] and write \(\mco_X(1)\) for the tautological twisting sheaf.
1. For any invertible sheaf \(\mcl\) on \(X\), there is a unique integer\(n\) such that\[ \mcl_y\cong\mco_{\PP^{r-1}_{\kappa(y)}}(n) \]for every \(y\in Y\).
For each \(y\), the fibre is \[ X_y\cong\PP^{r-1}_{\kappa(y)}. \] Since \(r-1\ge1\), \[ \Pic X_y\cong\ZZ \] with generator \(\mco_{X_y}(1)\). Thus there is a unique integer \(n(y)\) such that \[ \mcl_y\cong\mco_{X_y}(n(y)). \]
The sheaf \(\mcl\) is flat over \(Y\): locally it is \(\mco_X\), and \(p\) is flat. Hence constancy of Euler characteristic says that for every integer \(m\), the function \[ y\longmapsto \chi\bigl(X_y,\mcl_y(m)\bigr) \] is locally constant. The base \(Y\) is integral, hence connected, so this integer is independent of \(y\).
On the fibre, \[ \chi\bigl(X_y,\mcl_y(m)\bigr) = \chi\bigl(\PP^{r-1}_{\kappa(y)},\mco(m+n(y))\bigr) = \binom{m+n(y)+r-1}{r-1} \] as a polynomial in \(m\). Therefore the Hilbert polynomial \[ m\longmapsto\chi\bigl(X_y,\mcl_y(m)\bigr) \] is independent of \(y\). Distinct integers \(a\ne b\) give distinct translated polynomials \[ \binom{m+a+r-1}{r-1} \ne \binom{m+b+r-1}{r-1}; \] for example, their coefficients of \(m^{r-2}\) differ when \(r>2\), and for \(r=2\) they are the distinct linear polynomials \(m+a+1\) and \(m+b+1\). Thus \(n(y)\) is independent of \(y\). Denote its common value by \(n\).
2. Every invertible sheaf \(\mcl\) on \(X\) has the form\[ \mcl\cong p^*\mcn\tensor\mco_X(n) \]for some \(\mcn\in\Pic Y\) and \(n\in\ZZ\).
Let \(n\) be the integer from step 1 and put \[ \mcl_0=\mcl\tensor\mco_X(-n). \] Then for every \(y\in Y\), \[ (\mcl_0)_y\cong\mco_{X_y}. \]
The morphism \[ p:X\longrightarrow Y \] is flat and projective, and all of its fibres are projective spaces, hence integral. Apply Exercise III.12.4 to the two invertible sheaves \(\mcl_0\) and \(\mco_X\). There exists an invertible sheaf \(\mcn\) on \(Y\) such that \[ \mcl_0\cong p^*\mcn. \] Tensoring by \(\mco_X(n)\) gives \[ \mcl\cong p^*\mcn\tensor\mco_X(n). \]
3. Define\[ \Phi:\Pic Y\times\ZZ\longrightarrow\Pic X \]by\[ \Phi([\mcn],n) = [p^*\mcn\tensor\mco_X(n)]. \]Then \(\Phi\) is a surjective group homomorphism.
Pullback and tensor product give \[ p^*(\mcn_1\tensor\mcn_2) \cong p^*\mcn_1\tensor p^*\mcn_2, \] and \[ \mco_X(n_1)\tensor\mco_X(n_2) \cong \mco_X(n_1+n_2). \] Hence \(\Phi\) is a homomorphism. Its surjectivity is exactly step 2.
4. The homomorphism \(\Phi\) is injective.
Suppose \[ p^*\mcn\tensor\mco_X(n)\cong\mco_X. \] Restricting to any fibre gives \[ \mco_{\PP^{r-1}_{\kappa(y)}}(n)\cong \mco_{\PP^{r-1}_{\kappa(y)}}. \] Since \(r-1\ge1\) and \[ \Pic\PP^{r-1}_{\kappa(y)}\cong\ZZ[\mco(1)], \] we obtain \(n=0\). Thus \[ p^*\mcn\cong\mco_X. \]
For a projective bundle, \[ p_*\mco_X\cong\mco_Y. \] By the projection formula, \[ p_*p^*\mcn \cong \mcn\tensor p_*\mco_X \cong \mcn. \] Pushing forward \(p^*\mcn\cong\mco_X\) therefore gives \[ \mcn\cong\mco_Y. \] Hence the kernel of \(\Phi\) is trivial.
5. Consequently\[ \boxed{\Pic\PP(\mce)\cong\Pic Y\times\ZZ} \]when \(\operatorname{rank}\mce\ge2\).
Steps 3 and 4 show that \(\Phi\) is a bijective group homomorphism.
6. Q.E.D. for the corrected statement.
Step 5 proves the asserted formula under the necessary rank hypothesis, and the preceding erratum explains why the printed rank-one case cannot satisfy it.