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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.

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up vote 13 down vote accepted

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 $$ Pic(X)=\mathbb Z$$

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Here is an interesting link , internal to Wikipedia, about the trustworthiness of the answer of Wikipedia to the OP's question: – Georges Elencwajg Oct 5 '11 at 14:38
Thank you very much for your answer! Actually I supposed that was required a solution without the use of cohomological tools, but this one is really elegant. From the following link it is possible to find another one looking for the solution of Exercise 7.4. b). Also this other solution confirm that the correct conclusion is the one given by Georges. – Giovanni De Gaetano Oct 5 '11 at 14:52
@Student73: Dear Student, You can modify Georges's argument to avoid cohomology. Giving a line bundle on $X$ is the same as giving a line bundle on each of the copies of $\mathbb A^1$, with an identification between the two when restricted to their respective copies of $\mathbb A^1\setminus \{0\}$. Now, as you observed, the line bundles on $\mathbb A^1$ are trivial, and so you can choose a basis for the line bundles on each of the $\mathbb A^1$; this basis is determined up to multiplication by an element of $\mathcal O(\mathbb A^1)^{\times} = k^{\times}$. Having chosen these two bases, ... – Matt E Oct 5 '11 at 15:03
... the identification between the two line bundles restricted to $\mathbb A^1\setminus\{0\}$ is given by an element of $\mathcal O(\mathbb A^1\setminus \{0\}) = k^{\times} t^{\mathbb Z},$ where $t$ is the coordinate on $\mathbb A^1$. (This is a "change of basis matrix", which is just an invertible scalar in our situation, because we are dealing with line bundles.) So we find that $Pic(X) = k^{\times} \backslash ( k^{\times} t^{\mathbb Z} / k^{\times}$ (we have the gluing data on $\mathbb A^1\setminus \{0\}$, modulo the possible change of basis on each copy of $\mathbb P^1$), which equals – Matt E Oct 5 '11 at 15:09
$t^{\mathbb Z}$, just as Georges computed. This "double coset" approach to computing line bundles, or more generally vector bundles or principal $G$-bundles, on a space glued together out of two pieces is useful in lots of contexts. Regards, – Matt E Oct 5 '11 at 15:11

I will try to stay at the most elementary level possible.

Let $\mathcal L \in \mathrm{Pic}(X)$. Then $\mathcal L|_{U_1} \simeq \mathcal O_X|_{U_1}$ and $\mathcal L|_{U_2} \simeq \mathcal O_X|_{U_2}$ since $\mathrm{Pic}(\mathbb A^1_k)$ is trivial. Since $$ \Gamma(X,\mathcal O_X) \simeq \{ (f_1,f_2) \in k[x_1] \times k[x_2] \, | \, f_1(x) = f_2(x) \quad \forall x \in k \backslash \{0\} \} \simeq k[x], $$ we can inspire ourself from this definition and define invertible sheaves $\mathcal L_n$ for $n \in \mathbb Z$ and $U \subseteq X$ by "twisting" this glueing as follows : if $n \in \mathbb Z$, $$ \mathcal L_n(U) \overset{def}= \{ (s,t) \in \mathcal O_{U_1}(U \cap U_1) \times \mathcal O_{U_2}(U \cap U_2) \, | \, x^n (s|_{U \cap U_1 \cap U_2}) = t|_{U \cap U_1 \cap U_2} \}. $$ Note that $\mathcal L_n \otimes_{\mathcal O_X} \mathcal L_m \simeq \mathcal L_{m+n}$, so in particular $\mathcal L_n$ is invertible with inverse $\mathcal L_{-n}$.

Example. For $n \ge 0$ and $U \subseteq U_1$, we have $$ \mathcal L_n(U) = \{ (s,x^ns|_{U \cap U_2}) \in \mathcal O_X(U) \times \mathcal O_X(U \cap U_2) \} \simeq \mathcal O_X(U) $$ and these isomorphisms commute with restriction so we have $\mathcal L_n|_{U_1} \simeq \mathcal O_X|_{U_1}$. Similarly, for $U \subseteq U_2$ we have $$ \mathcal L_n(U) = \{ (s|_{U \cap U_1},x^n s) \in \mathcal O_X(U \cap U_1) \times \mathcal O_X(U) \} \simeq \mathcal O_X(U) $$ and this time the isomorphism of $\mathcal O_X(U)$-modules projects on the second component instead of the first, which explains why $\mathcal L_n$ does not glue to $\mathcal O_X$ ; these sheaf isomorphisms do not agree on overlaps, differing by a factor of $x^n$. Since $U \cap U_1$ does not contain the origin of $U_1$, this allows the first component $s|_{U \cap U_1}$ to have a pole of order up to $n$ at zero when $U \subseteq U_2$.

Now this proves that we have a group homomorphism $\mathbb Z \to \mathrm{Pic}(X)$ (in a very explicit manner, which I like) but it does not prove that this is the whole of the Picard group, i.e. that our map is an isomorphism. For this we use Cartier divisors, which is a bit less pedagogical for the Picard group but more efficient.

Note that since $X$ is integral, the sheaf $\mathcal K^{\times}$ is constant and equal to $k(x)$. A Cartier divisor on $X$ can be described by the following data : a collection $\{(V_i,f_i)\}$ where the $V_i$ cover $X$ and the $f_i$ are such that on $V_i \cap V_j$, the section $f_i/f_j$ lies in $\Gamma(V_i \cap V_j, \mathcal O_X^{\times})$. We can restrict Cartier divisors to $U_1$ and $U_2$, and since they each have trivial Picard group, their Cartier divisors are all principal, that is, Cartier divisors on $X$ are given by pairs $(f_1,f_2)$ such that $f_i \in \Gamma(U_i,\mathcal K^{\times})$ and $f_1/f_2 \in \Gamma(U_1 \cap U_2, \mathcal O^{\times})$. Finally, the Cartier class group is the group of Cartier divisors modulo principal Cartier divisors.

General Cartier divisors go as follows : we are given a pair $(f_1,f_2) \in (k(x)^{\times})^2$ such that $$ f_1/f_2 \in k[x]_x^{\times} = \{ ax^n \, | \, a \in k, \quad n \in \mathbb Z \}. $$ A Cartier divisor given by the pair $(f_1,f_2)$ is principal if and only if $f_2 = a f_1$ for some $a \in k^{\times}$. To see this, it is clear that the Cartier divisor $(f,f)$ is principal ; so it remains to see that the Cartier divisor $(1,a)$ is principal for $a \in k^{\times}$. But this is clear since it generates the subsheaf $\mathcal O_X$ of $\mathcal K$ (see Hartshorne, Chapter II, Proposition 6.13 (c)).

This is something that confused me for a long time ; although a Cartier divisor can be described by a collection $\{(U_i,f_i)\}$, it is not characterized by it, even if we fix the cover we are defining it on ; changing one $f_i$ by an element of $\mathcal O_X(U_i)^{\times}$ does not change the Cartier divisor since the stalk of the corresponding global section of $\mathcal K^{\times}/\mathcal O^{\times}$ remains unchanged. So the Cartier divisor given by $(1,1)$ is in fact the same Cartier divisor given by $(1,a)$.

So letting $H \subseteq (k(x)^{\times})^2$ be the subgroup of all $(f_1,f_2)$ satisfying the above Cartier condition, and modding out $H$ by the subgroup of all principal divisors, we can write any element of the quotient in the form $(1,x^n)$ for some $n \in \mathbb Z$.

We deduce $$ \mathrm{Pic}(X) \simeq \mathbb Z. $$ Also note that the group isomorphism $\mathcal{CaCl}(X) \to \mathrm{Pic}(X)$ sends the Cartier divisor corresponding to $(1,x^n)$ to the invertible sheaf $\mathcal L_{-n}$ since $\mathcal L_{-n}$ is generated by $1$ on $U_1$ and by $x^{-n}$ on $U_2$ (when we see $\mathcal L_{-n}$ as an invertible subsheaf of $\mathcal K$ using the embedding $\mathcal L_{-n} \to \mathcal L_{-n} \otimes_{\mathcal O_X} \mathcal K \simeq \mathcal K$).

Hope that helps,

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