Let $X$ be the affine line with double origin over a field $k$. It is the scheme obtained gluing two copies of the affine line $\mathbb{A}^1_k$ along the open sets $U_1 = U_2 =\mathbb{A}^1_k – (x)$, where, with abuse of notation, $(x)$ is the point associated to the maximal ideal of $k[x]$ generated by $x$. It is the construction of Example 2.3.6 of Chapter II of Hartshorne´s Algebraic Geometry.
A part of Exercise 7.4 of the same chapter of the same book asks to find the Picard Group of this scheme $X$, I don´t know how to find it.
I made a few observations about the matter, the first one is that $Pic(\mathbb{A}^1_k)=0$, so the untriviality of $Pic(X)$ is concentrated on the double point. The second one is that, against my intuition, $X$ is an integral scheme. Indeed it is clearly irreducible, and the existence of a nilpotent in $\mathcal{O}_X(U)$ for an open set $U$ of $X$ containing at least one of the two origins implies the existence of a nilpotent in $\mathcal{O}_{\mathbb{A}^1_k}(V)$, where $V$ is the preimage of $U$ in the affine line to which belongs the origin contained in $U$.
Using the integrality of $X$ and Proposition 6.15 of Chapter II of Hartshorne´s Algebraic Geometry we deduce that $Pic(X)$ is isomorphic to $\mathcal{CaCl}(X)$, i.e. the group of Cartier divisors on $X$ modulo linear equivalence. But I don´t know how to go further.
A last notification is that Wikipedia states the result, and it is $Pic(X)\simeq \mathbb{Z}\times k^*$.
Thank you in advance for your time.
Best Answer
Let's exploit the isomorphism $Pic(X)=H^1(X,\mathcal O^\ast)$ to attack the question.
To compute $H^1(X,\mathcal O^\ast)$ we'll use Cech cohomology and the open covering of $\mathcal U$ of $X$ by the two obvious open subsets $U_1, U_2\subset X$ isomorphic to $\mathbb A^1_k$.]
The crucial point is that this covering is acyclic for $\mathcal O^\ast_X$, because
1) $H^1(U_i,\mathcal O^\ast)=H^1(\mathbb A^1_k,\mathcal O^\ast)=0$ because $Pic(\mathbb A^1_k)=0$
2) $H^1(U_{12},\mathcal O^\ast_X)=0 $ because $U_{12}$ is isomorphic to $\mathbb A^1_k \setminus 0$ , which also has zero Picard group.
3) $H^p(U_i,\mathcal O^\ast_X)=H^p(U_{12},\mathcal O^\ast_X)=0$ for $p\geq 2 $ , because cohomology vanishes above the Krull dimension of a space.
Hence Leray's theorem says that $ H^1(X,\mathcal O^\ast_X)= \check H^1(\mathcal U,\mathcal O^\ast_X)$
So the required group $\check H^1(\mathcal U,\mathcal O^\ast_X)$ is the quotient of the cocycle group $\mathcal O^\ast_X(U_{12})$ by the coboundary subgroup $B$.
Final we remark that $\mathcal O^\ast_X(U_{12})$ consists of the rational functions $g_{12}=az^n \; (a\in k^\ast, n\in\mathbb Z)$ and $B$ of the quotients $g_2/g_1 \; (g_1,g_2\in k^\ast)$.
Conclusion $$ Pic(X)=\mathbb Z$$