How to prove $\operatorname{Pic}(\mathbb{P}_X^n)\cong\operatorname{Pic}(X)\times\mathbb{Z}$? Let $X$ be a Noetherian regular scheme. Then how can one prove $\operatorname{Pic}(\mathbb{P}_X^n)\cong\operatorname{Pic}(X)\times\mathbb{Z}$? I want to use this specific case for the more general case where one replaces $\mathbb{P}_X^n$ with $\mathbb{P}(\mathcal{E})$ and $\mathcal{E}$ a locally free sheaf of rank $n+1$, where the result is still true. So the proof shouldn't use this more general result. Is there a simpler proof for $\mathbb{P}_X^n$ than for $\mathbb{P}(\mathcal{E})$?
 A: Let me first note that you should require $X\newcommand{\Pic}{\operatorname{Pic}} \newcommand{\Spec}{\operatorname{Spec}} \newcommand{\PP}{\mathbb{P}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\Cl}{\operatorname{Cl}}$ connected - if not, you can get something silly like $X=\Spec k \sqcup \Spec k$ with trivial Picard group and $\PP^1_X=\PP^1_k \sqcup \PP^1_k$, which has Picard group $\ZZ^2$.

There's an obvious choice for the map $\Pic(X)\times\ZZ\to \Pic(\PP^n_X)$: send $$(\mathcal{L},d)\mapsto (p^*\mathcal{L})\otimes(q^*\mathcal{O}(d))$$ where $p:\PP^n_X\to X$ and $q:\PP^n_X\to \PP^n$ are the projections, so we show this map is injective and surjective.
Injectivity: Suppose $p^*\mathcal{L}\otimes q^*\mathcal{O}(d)\cong \mathcal{O}_{\PP^n_X}$. By a relative version of the fact that $\mathcal{O}_{\Bbb P^n_A}(d)$ is the $d$-graded piece of $A[x_0,\cdots,x_n]$, we see that $p_*(p^*\mathcal{L}\otimes q^*\mathcal{O}(d))\cong \mathcal{O}_X$, so by the projection formula we have that $\mathcal{L}\otimes p_*q^*\mathcal{O}(d)\cong\mathcal{O}_X$, so $d=0$ and $\mathcal{L}\cong\mathcal{O}_X$.
Surjectivity: Pick an affine open cover $\{U_i=\Spec A_i\}$ of $X$ with each $A_i$ an integral domain. Then each $U_i$ and hence each $U_i\times\PP^n$ is regular, noetherian, integral, and separated, so $$\Pic (U_i\times\PP^n) \cong \Cl (U_i\times\PP^n) \cong (\Cl U_i) \times\ZZ \cong (\Pic U_i)\times\ZZ$$ by the results of Hartshorne chapter II section 6, where the map is the same as we've picked above. It remains to check that these isomorphisms patch together correctly.
So let $U_i,U_j$ be two of the opens from above. From the previous isomorphism of Picard groups, we see that the restriction of our line bundle $\mathcal{L}$ on $X\times\PP^n$ is of the form $p_i^*\mathcal{L}_i\otimes q_j^*\mathcal{O}(n_i)$ on $U_i\times\PP^n$ and $p_j^*\mathcal{L}_j\otimes q_j^*\mathcal{O}(n_j)$ on $U_j\times\PP^n$. Since the restriction of both of these to the intersection $U_i\times\PP^n\cap U_j\times\PP^n$ is the same, we see again from the first result in the proof of injectivity that $n_i=n_j$. Untwisting by $n_i=n_j$ along $U_i\times\PP^n$ and $U_j\times\PP^n$, we see that we get a line bundle trivial in the $\PP^n$ direction on $U_i\times\PP^n$ and $U_j\times\PP^n$. The claim is that this comes from a line bundle on $X$: after taking the gluing data for $\mathcal{L}$, applying the above untwisting and the projection formula gives us that it is exactly the gluing data for a line bundle on $X$ which pulls back correctly. So we've shown surjectivity.

Note that this same proof works for $\PP(\mathcal{E})$ by requiring $\mathcal{E}$ to be trivial along the $U_i$, which is always possible.
