Picard Group of Projective Bundle over an Integral scheme

Question

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.

Answer 1

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

$$ CH^\ast(\mathbb P(\mathcal E) )= CH^\ast (Y)[\xi]/ < \xi^n +c_1 (p^\ast \mathcal E)\xi^{n-1} +\cdots+c_n (p^\ast \mathcal E)> $$ 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 semi-continuity theorem (cf. Hartshorne Chapter III, §12) rather than Chow groups.

Answer 2

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 Riemann-Roch). 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.

Answer 3

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 $X-U=:Z$, 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: $$\mathbb{Z} \rightarrow \mbox{Pic}(X) \rightarrow \mbox{Pic}(X-Z)\rightarrow 0$$ and $$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(\mathbb{P}(E|_{X-Z}))\rightarrow 0$$ 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 $$\mathbb{Z} \rightarrow \mbox{Pic}(X)\oplus\mathbb{Z} \rightarrow \mbox{Pic}(X-Z)\oplus\mathbb{Z}\rightarrow 0$$ And for the second sequence, since $E|_{X-Z}$ 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 $$\mathbb{Z} \rightarrow \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(X-Z)\oplus\mathbb{Z}\rightarrow 0$$ 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 $p: E-X \rightarrow \mathbb{P}(E)$ gives us another pullback $$p^{*}: \mbox{Pic}(\mathbb{P}(E))\rightarrow \mbox{Pic}(E-X).$$ But we may of course assume rank of $E$ is $\geq 2$, then $X$ has codimension $\geq 2$ in $E$, which implies $$\mbox{Pic}(E-X)=\mbox{Pic}(E).$$ 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.