Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

1 Answer 1

up vote 10 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 $$ Pic(X)=\mathbb Z$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.