The Picard Group of the Affine line with double origin

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 $Pic(X)\simeq \mathbb{Z}\times k^*$.

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

• Here is an interesting link , internal to Wikipedia, about the trustworthiness of the answer of Wikipedia to the OP's question: en.wikipedia.org/wiki/Talk%3APicard_group Commented Oct 5, 2011 at 14:38
• @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, ... Commented Oct 5, 2011 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 Commented Oct 5, 2011 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, Commented Oct 5, 2011 at 15:11
• +1 for Matt's comments, since this is really geometry and in this case cohomology and Leray's theorem are just overkill. By the way, using Smith normal form one sees that $\mathrm{Vect}_n(X)=\mathrm{GL}_n(k[t]) \backslash \mathrm{GL}_n(k[t,t^{-1}]) / \mathrm{GL}_n(k[t]) = \mathbb{Z}$ and therefore $\mathrm{Vect}(X)=\mathbb{N}[t,t^{-1}]$ (Laurent series semiring). Commented Aug 23, 2013 at 7:14

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,

• Thanks for the explanations. I wonder How much freedom $(f_i)$ can have such that the Cartier divisor will be preserved. As you pointed out, $f_i$ can be changed by a unite of $\mathcal{O}_X(U_i)$. But if $f_i$ is changed by a non unite, is it necessary the corresponding Cartier divisor is changed as well? Thanks. Commented May 11, 2017 at 1:24
• @Shuhang : if you read through my argument carefully, you'll realize that I produced an isomorphism $$\mathrm{Pic}(X) \simeq (k(x)^{\times})^2/ H, \quad H = \{ (f_1,f_2) \in (k(x)^{\times})^2 \, | \, f_2 \equiv f_1 \pmod{\Gamma(U_1 \cap U_2, \mathcal O_X^{\times}) } \}.$$ Note that the group operation in $H$ (as in $(k(x)^{\times})^2$) is multiplication. Also note that $\Gamma(U_1 \cap U_2, \mathcal O_X^{\times}) = k[x]_x^{\times} = \{ ax^n \, | \, a \in k^{\times}, n \in \mathbb Z \}$. Commented May 14, 2017 at 8:20
• @Shuhang : Since it seemed to interest you, I added a description of $\mathrm{Pic}(X)$ in terms of the sections on the standard open affine cover of the scheme. Feel free to ask me any questions. Commented May 14, 2017 at 9:28
• Thank You for the explanation Commented May 15, 2017 at 22:04

Note that we can easily cover $X$ with two affines, namely $U_1$ and $U_2$. Furthermore, since any line bundle will be trivial on $U_1$ and $U_2$ means that any line bundle on $X$ will trivialize when restricted to these two open subsets.

Now the problem reduces to figuring out exactly what the invertible functions on $U_1 \cap U_2 = \mathbb{A}^1 -\{0\}$ are since these will be our only options for transition functions. I like to think of invertible functions on $\mathbb{A}^1 - \{0\}$ as all function which have neither pole nor zero anywhere on $\mathbb{A}^1$ except for possibly at zero.

A quick thought will lead you to realize that such functions must be of the form $k^{\times} t^{\mathbb{Z}}$ (I stole this notation from Matt E.) Now we have a whole lot of choices for transition functions but how can we use these to determine how many isomorphism classes of line bundles we have?

Use the following fact which I motivated from cohomology: Two invertible functions $s_1, s_2 \in U_1 \cap U_2$ define the same line bundle on $X$ if $$s_1 s_2^{-1} = r_1 \vert_{U_1 \cap U_2} r_2^{-1} \vert_{U_1 \cap U_s}$$

where $r_1 \in \mathcal{O}^{\times}(U_1)=k^{\times}$ and $r_2 \in \mathcal{O}^{\times}(U_2)=k^{\times}$

This immediately implies that two invertible functions on $U_1 \cap U_2$ define the same line bundle if and only if they differ by a nonzero element of $k$.

Thus $$Pic(X) = k^{\times} t^{\mathbb{Z}} / k^{\times} = t^{\mathbb{Z}} \cong \mathbb{Z}$$