Let $X=\mathbb P(\mathcal E)$, where $\mathcal E$ is a locally free sheaf of rank $n+1$ on $Y$, an integral scheme of finite type over an algebraically closed field $k$. I'm trying to show that $\text{Pic }X\cong \text{Pic }Y\times \mathbb Z$. The only small point I'm stuck on is showing that every invertible sheaf on $X$ is of this form. I consider an invertible sheaf on $X$, $\mathcal M$, and it's restriction $\mathcal M_y$ to the fiber $X_y$ over a point $y$. Since this is an invertible sheaf on $\mathbb P^n$ we get that it must be $\mathcal O_{\mathbb P^n}(m)$ for some $m$. So I consider $\mathcal M\otimes \mathcal O_X (m)$, where the second term in the tensor product comes from the natural invertible sheaf $\mathcal O_X(1)$ on the projective bundle. My only question is how do I know that on another fiber, say $X_{y'}$, $\mathcal M\otimes \mathcal O_X(m)$ will be isomorphic to $\mathcal O_{X_{y'}}$ like it is on the fiber over $y$. Once I have this I can use something else I've shown to finish up.

$\begingroup$ It is effective and of degree $0$ on every fibre, so the restriction is trivial for every $y$. $\endgroup$– Francesco PolizziApr 8, 2011 at 12:40

$\begingroup$ How do you see that it's effective and of degree 0 on every fibre? $\endgroup$– HNuerApr 8, 2011 at 12:43

1$\begingroup$ The degree of $\mathcal{L}:=\mathcal{M} \otimes \mathcal{O}_X(m)$ restricted to a fibre $F$ is given by the square root of the intersection number $\mathcal{L} \cdot \mathcal{L} \cdot F$. This is clearly independent on the fibre, since all the fibres are algebraically equivalent. It follows that $\mathcal{L}$ has degree $0$ when restricted on each fibre. Since the fibre is a projective space, the restriction must be trivial. $\endgroup$– Francesco PolizziApr 8, 2011 at 12:53

$\begingroup$ Doesn't effectiveness then follow from the degree being 0? Is there a way of seeing this wihtout the notion of degree? Thanks for your help. $\endgroup$– HNuerApr 8, 2011 at 12:56

$\begingroup$ Yes it follows from the degree being 0. It seems to me that the degree is the natural way to treat this kind of problems $\endgroup$– Francesco PolizziApr 8, 2011 at 13:04
3 Answers
Dear HNuer, a fundamental theorem on Chow groups describes the relation between the Chow ring $CH^\ast (\mathbb P(\mathcal E))$ of $X=\mathbb P(\mathcal E)$ and that $CH^\ast (Y)$ of $Y$ when $Y$ is a regular variety over a not necessarily algebraically closed field.
If we call $p:\mathbb P(\mathcal E) \to Y $ the projection and $\xi$ the relative hyperplane bundle $\mathcal O_{\mathbb P(\mathcal E)}(1)$, we have
$$ CH^\ast(\mathbb P(\mathcal E) )= CH^\ast (Y)[\xi]/ < \xi^n +c_1 (p^\ast \mathcal E)\xi^{n1} +\cdots+c_n (p^\ast \mathcal E)> $$ In particular $CH^1(\mathbb P(\mathcal E) )=p^\ast CH^1(Y)\oplus \mathbb Z \xi. $ (This is true even if $Y$ is not regular)
If you remember that locally factorial varieties (for example regular or smooth varieties) satisfy $Pic(P)=CH^1 (P)$ , your formula is proved under this hypothesis of local factoriality.
Edit: As the OP remarks in his comments below, the formula $Pic(\mathbb P(\mathcal E) )=p^\ast Pic(Y)\oplus \mathbb Z \xi $ is also true for any integral variety $Y$ over an algebraically closed field, locally factorial or not. The tool is then Grauert's semicontinuity theorem (cf. Hartshorne Chapter III, §12) rather than Chow groups.

$\begingroup$ I don't yet know too much above the Chow ring, so I didn't understand your answer so much. Either way, I am in fact trying to prove this for a general integral scheme, and it seems to be true there, so I'm not assuming anything about the singularities of my scheme. $\endgroup$– HNuerApr 9, 2011 at 17:36

$\begingroup$ There is a misunderstanding here. My claim is that the formula $Pic(\mathbb P(\mathcal E) )=p^\ast Pic(Y)\oplus \mathbb Z \xi $ is proved above only under the local factoriality hypothesis. In particular if a bundle on $\mathbb P(\mathcal E)$ is trivial on all fibers $p^{1}(y)$, is it of the form $p^{\ast} \mathcal L$ for some line bundle $ \mathcal L$ on $Y$, if nothing is assumed on the singularities of $Y$? This question has not been addressed in the comments nor in your answer and since in the original question you seem to claim that the formula is true, could you plese explain why ? $\endgroup$ Apr 10, 2011 at 21:53

$\begingroup$ I understand your question now. In my answer I just showed why if we take the line bundle $M$ on $\mathbb P(\mathcal E)$ it is isomorphic on each fiber to the same $\mathcal O_{\mathbb P^n}(m)$ and that this $m$ doesn't change. Once we have that, we in the situation of Hartshorne exercise III,12.4 with a flat projective morphism $\pi:\mathbb P(\mathcal E)\rightarrow Y$ and two invertible sheaves $\mathcal M$ and $\mathcal O_{\mathbb P(\mathcal E)}(m)$ which are isormorphis on each fiber. By that exercise (which is essentially Grauert's theorem), we get a line bundle $\mathcal N$ with $\endgroup$– HNuerApr 11, 2011 at 7:04

$\begingroup$ , the line bundle $\mathcal N$ is on $Y$, $\mathcal M\cong \mathcal O_{\mathbb P(\mathcal E)}(m)\otimes \pi^* \mathcal N$ $\endgroup$– HNuerApr 11, 2011 at 7:06

$\begingroup$ So certainly every line bundle can be written this way, and injectivity follows from the properties of $\mathcal O_{\mathbb P(\mathcal E)}$ and the projection formula for locally free sheaves. $\endgroup$– HNuerApr 11, 2011 at 7:07
Thanks to Piotr Achinger for the idea to consider the euler characteristic. I was looking for an answer that doesn't use fancy machinery beyond what's presented in the main text in Hartshorne (so no generalized RiemannRoch). Here is one based on his suggestion:
Denote by $\mathcal F$ the line bundle $\mathcal M\otimes \mathcal O_X(m)$ with notation as above. Then we have that on the fiber above our point $y$, $\mathcal F_y=\mathcal O_{X_y}$. Now since $Y$ is an integral scheme, it's connected, and since the Euler characteristic is constant in this case, we see that $\chi(\mathcal F)(y')$ is the constant function with value 1 since it takes that value at the point $y$. But since on $\mathbb P^n$ lines bundles have no cohomology between $H^0$ and $H^n$, we get that $1=\chi(\mathcal F)(y')=h^0(y',\mathcal F)+(1)^n h^n(y',\mathcal F)$. But this implies that on each fiber $\mathcal F_y'$ is the trivial line bundle or the canonical bundle (if $n$ is even, otherwise we get the result immediately since then the Euler characteristic would be 1) since in every othercase either both $h^0$ and $h^n$ vanish, or just $h^n$ vanishes but then $h^0$ is too large.
Now to prove that we in fact always get the trivial line bundle on fibers, we consider $h^0(y',\mathcal F)$. By semicontinuity we get that since the only values possibly taken are 0 and 1, the set $S$ upon which 0 is achieved by $h^0(y',\mathcal F)$ is open (being the complement of the closed set when this function is $\geq 1$). Now considering everything above with $\mathcal F^{1}$ instead, we get that the set upon which 0 is acheieved for $h^0(y',\mathcal F^{1})$ is also open. But this must be the complement of $S$. So $S$ is both open and closed in a connected space, and thus it's either empty or the entire space. It can't be the entire space since our point $y$ is not in it. Hence it's empty and $h^0(y',\mathcal F)=1$ everywhere. This gives us that $\mathcal F_y=\mathcal O_{X_y}$ on every fiber.

1$\begingroup$ See also Mumford, Fogarty, Kirwan, Geometric Invariant Theory, 0, Sec. 5, b) and Vakil, Foundations of Algebraic Geometry, Exercise 28.1.K. $\endgroup$ Nov 22, 2017 at 21:56
I think under some assumption (your conditions might already suffice) you may use excision.
Pick an open set $U$ on $X$ such that the vector bundle $E$ is trivial, denote $XU=:Z$, and write $Z$ to be $Z_{1}$, which is a divisor, and some higher codimensional stuff, say $Z_{\geq 2}$. Then using excision, higher codimensional terms may be ignored, namely we may assume $Z_{\geq 2}=0$. Also by induction we may assume $Z=Z_{1}$ is a prime divisor. Then we apply excision to $(X,Z)$ get two exact sequences: $$\mathbb{Z} \rightarrow \mbox{Pic}(X) \rightarrow \mbox{Pic}(XZ)\rightarrow 0$$ and $$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(\mathbb{P}(E_{XZ}))\rightarrow 0$$ Now we can modify the first sequence by adding $\mathbb{Z}$ terms to the second and third term with the identity map between them, then the sequence is still exact, namely we have $$\mathbb{Z} \rightarrow \mbox{Pic}(X)\oplus\mathbb{Z} \rightarrow \mbox{Pic}(XZ)\oplus\mathbb{Z}\rightarrow 0$$ And for the second sequence, since $E_{XZ}$ is trivial, by an exercise on Hartshorne, $\mbox{Pic}(Y\times \mathbb{P}^{n})=\mbox{Pic}(Y)\oplus \mathbb{Z}$, apply this to the second sequence we have $$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(XZ)\oplus\mathbb{Z}\rightarrow 0$$ Now the first and third terms of these two sequences are equal, and there is a map on the second position, denoted, say, $\pi^{*}\oplus \phi$, where $\pi^{*}$ is the pull back of $\pi: \mathbb{P}(E)\rightarrow X$, and $\phi$ sends $1$ to $\mathcal{O}(1)$.
Now then the result follows from five lemma if we can show $\pi^{*}$ is injective, since $\phi$ is already injective. By abusing of notation we denote the total space of vector bundle $E$ by $E$ and $i: X\rightarrow E$ be the inclusion, we get pullback $$i^{*}: \mbox{Pic}(E)\rightarrow \mbox{Pic}(X).$$ Also, the projection $p: EX \rightarrow \mathbb{P}(E)$ gives us another pullback $$p^{*}: \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(EX).$$ But we may of course assume rank of $E$ is $\geq 2$, then $X$ has codimension $\geq 2$ in $E$, which implies $$\mbox{Pic}(EX)=\mbox{Pic}(E).$$ These maps concatenate to a map $\mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(X)$, which is easily seen to be a section of $\pi^{*}$. Hence $\pi^{*}$ is injective.

$\begingroup$ You seem to be assuming that $X$ is smooth (or at least locally factorial), which is definitely unnecessary. $\endgroup$– abxOct 22, 2019 at 20:04

$\begingroup$ I think you are right, I'm using Pic=Cl. Actually I encountered this when doing Hartshorne exercise 7.9, and indeed he assumed smoothness, but I don't know to what extent this condition can be weakened so that excision in Picard groups can be used, so that's why I said ''under some good conditions''. $\endgroup$– Y_qOct 22, 2019 at 22:31