TITLE: Picard group of scheme over DVR
QUESTION [9 upvotes]: Let $A$ be a DVR and let $X/A$ be a smooth, proper scheme with geometrically integral fibers. Is there an easy way to see that the Picard group of $X$ is isomorphic to the Picard group of the generic fiber $X_\eta$ of $X$?

REPLY [9 votes]: This is too long to be a comment to Matt Emerton's answer. I think his conclusion is correct. Namely, if $X$ is smooth and proper over $A$, then $\mathrm{Pic}(X)\to \mathrm{Pic}(X_{\eta})$ is an isomorphism (whenever the generic fiber is geometrically connected or not). As explained by Matt, one can suppose $X$ is connected and we have only to show the injectivity. Clearly the kernel is generated by the irreducible components of the special fiber $X_s$. So it is enough to show that each of these irreducible components is a principal (Cartier) divisor. 
Frist consider $A'=H^0(X, \mathcal O_X)$. It is finite and flat over $A$. As $X_s=X\otimes_{A'} (A'/\pi A')$ where $\pi$ is a uniformizing element of $A$, is smooth over $A/\pi A$, this force $A'/\pi A'$ to be separable over $A/\pi A$, hence $A'$ is étale over $A$. 
The scheme $X$ is also a smooth proper $A'$-scheme with geometrically connected generic fiber. Let $s_1,\dots, s_n$ be the points of Spec$(A')$. Then $A'/\pi A'=\oplus_i k(s_i)$ and $X_s$ is just the disjoint union of the $X_{s_i}$. By Zariski, $X_{s_i}$ is connected, smooth, hence irreducible. Let $\pi_i\in A'$ be a uniformizing element at $s_i$ and a unit at $s_j$ for all $j\ne i$. Then the divisor div$(\pi_i)$ of $\pi_i$ on $X$ is $X_{s_i}$. 
In Laurent's example, each copy of $\mathbb P^1_k$ is actually a principal divisor. So there is no more divisors on $\mathbb P^1_{A'}$ than on $\mathbb P^1_{\mathrm{Frac}(A')}$. Hope I don't miss something.