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. 



Picard Group of Projective Bundle over an Integral scheme
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.


