Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What's the strategy for computing the Picard group of a variety with more than one irreducible components?

For instance, consider the simple case where $X$ has two components $C$ and $D$, meeting transversely at one point. Then it seems that $\text{Pic}(X)=\text{Pic}(C)\times\text{Pic}(D),$ but I don't know how to prove it.


Edit: I'd like to see a formal proof (or a reference), i.e. using cohomology etc., instead of using "gluing", especially when we are not gluing along open intersections. The map $C\coprod D\to X$ is not flat, so there's no fppf gluing either.

share|improve this question
It follows from the description $Pic(X)=H^1(X,O_X^*)$ that there is a surjection $Pic(X)\to \bigoplus Pic(X_i)$ where $X_1,\ldots,X_r$ are the irreducible components of $X$ (cf. ex. III.5.8 in Hartshorne). –  J.C. Ottem Mar 2 '11 at 17:50
@JC: this is OK as long as $\dim X=1$. In higher dimensions you have to worry about the restrictions to the intersections agreeing. –  Sándor Kovács Mar 2 '11 at 17:56
Yes, for some reason I thought the OP considered curves $C$ and $D$. –  J.C. Ottem Mar 2 '11 at 18:02
@shanghao: in the answer you accepted, the map $\mathrm{Pic}(C) \times \mathrm{Pic}(D)\to H^1(X, F)$ measures whether a pair of line bundles "glue" or not. If you want a description of $\mathrm{Pic}(X)$ in terms of $\mathrm{Pic}(C)$ and $\mathrm{Pic}(D)$ you can't escape dealing with this somewhere. Of course, you can dig your head into the sand and pretend that it is not there. It works well for ostriches. Cheers. –  Sándor Kovács Mar 2 '11 at 21:30

4 Answers 4

up vote 12 down vote accepted

I will only consider the case of connected projective (this is not really necessary) curves $X, C, D$. over an algebraically closed field $k$. The canonical injection $O_X\to O_C\times O_D$ induces an exact sequence of sheaves on $X$ $$ 1 \to O_X^* \to O_C^* \times O_D^* \to F \to 1 $$ where $F$ is a skyscrapper sheaf supported at the intersection points of $C$ and $D$. Passing to cohomology, we get $$ 1 \to k^* \to k^* \times k^* \to F(X) \to \mathrm{Pic}(X)\to \mathrm{Pic}(C) \times \mathrm{Pic}(D)\to H^1(X, F)=0.$$ When $C, D$ intersect transversally at a single point (ordinary double point), a local computation shows that $k^* \times k^*\to F(X)$ is surjective, and you get your isomorphism.

As Steven said, this really depends on how $C$ and $D$ intersect (in fact, when $C$ and $D$ are smooth, your isomorphism implies that $C$ and $D$ intersect transversally at a single point; non transversal intersection point can give additive subgroup in $\mathrm{Pic}(X)$ and more transversally intersection points give subtori in $\mathrm{Pic}(X)$). Also in higher dimension $H^1(X, F)$ may not vanish, and the above methode does not work.

The general picture for proper curves can be found in Bosch, Lütkebohmert and Raynaud, Néron Models, Chap. 9., §2. They make a ''dévissage'' of $\mathrm{Pic}^0(X)$.

[EDIT] Let me rewrite Sándor's nice interpretation in cohomological terms. It will make explicit the sheaf $F$ above and gives a better understanding of what is going on, and in any dimension. Denote by $E=C\cap D$ the closed subscheme defined by the ideal $J_C+J_D$. Then we have an exact sequence of sheave on $X$: $$ 1 \to O_X^* \to O_C^* \times O_D^* \to O^*_{E} \to 1 $$ in the middle, the map is $(a, b)\mapsto a|_E (b|_E)^{-1}$. The exactness is checked locally. Passing to cohomology, we get $$ O(C)^\star\times O(D)^\star\to O(E)^* \to \mathrm{Pic}(X) \to \mathrm{Pic}(C)\times \mathrm{Pic}(D)\to \mathrm{Pic}(E)$$ and the last map is $(L, H)\mapsto L_{|E}\otimes (H_{|E})^{-1}$, therefore the exact sequence becomes

$$ O(C)^\star\times O(D)^\star\to O(E)^\star \to \mathrm{Pic}(X) \to \mathrm{Pic}(C)\times_{\mathrm{Pic}(E)} \mathrm{Pic}(D)\to 1.$$ So Sándor's map $\mathrm{Pic}(X) \to \mathrm{Pic}(C)\times_{\mathrm{Pic}(E)} \mathrm{Pic}(D)$ is always surjective. It is injective if and only if $O(C)^\star\times O(D)^\star\to O(E)^\star$ is surjective. This is not always the case (consider two irreducible curves meeting at more than one point or meeting at a single point but not transversally which implies that $E$ is a non-reduced point), but is true if for instance $O(E)=k$ (e.g. $E$ is geometrically connected, geometrically reduced and proper).

[EDIT 2] Of course, in all this answer, $X$ is supposed to be reduced. Otherwise $O_X\to O_C\times O_D$ (and that one with invertible functions) would not be necessarily injective. If $X$ is not reduced, there is a dévissage from $\mathrm{Pic}(X)$ to $\mathrm{Pic}(X_{\mathrm{red}})$ in Bosch & al, op. cit.

share|improve this answer
+1 . –  Qfwfq Mar 2 '11 at 18:57
So... If we let $X=Spec(k[x,y]/xy(x-y))$, $C=Spec(k[x,y]/(xy))$ and $D=Spec(k[x,y]/(x-y)$, your version of Sandor won't work since $E=Spec(k[x,y]/(xy,x-y))$ has units that don't lift, so your short exact sequence is not exact on the right. But if we redefine $E$ to be $Spec(k[x,y]/(x,y)$ then it seems like the argument works perfectly and shows that $Pic(X)$ is trivial. Is that right or am I missing something? –  Steven Landsburg Mar 3 '11 at 3:47
The sequence of sheaves is exact on the right because at any point $p$, $O_{C,p}^{\star}\to O_{E,p}^{\star}$ is already surjective (any lifting of an invertible element in the quotient is invertible because we are dealing with local rings). As these sheaves are not quasi-coherent, the surjectivity does'nt extend to $X$. In your example, $\mathrm{Pic}(X)$ is non-trivial: embed $X$ in a projective cubic curve $P$, then $\mathrm{Pic}(X)$ is isomorphic to $\mathrm{Pic}^0(P)$ (the power $0$ means degree $0$ on every irreducible component), and the latter is known to be the additive group $k$. –  Qing Liu Mar 3 '11 at 8:58
Continued: This coincides with the cokernel of $k^{\star}\times k^{\star}=O(C)^{\star}\times O(D)^{\star}\to O(E)^{\star}=k^*+k{\bar{y}}$. I think this is correct. –  Qing Liu Mar 3 '11 at 9:03
Qing: I made two silly mistakes. One (as you point out) was thinking I didn't have exactness on the right and the other was thinking that I could change $E$ and still retain exactness on the left. The latter is what led to the false computation $Pic(X)=0$, which I felt sure had to be wrong. Thanks for bearing with me. –  Steven Landsburg Mar 3 '11 at 15:55

It is certainly not always true that it is the product. Consider this: you have two restriction maps ${\rm Pic}(C)\to {\rm Pic} (C\cap D)$ and ${\rm Pic}(D)\to {\rm Pic} (C\cap D)$. In order for the two line bundles you pick on $C$ and $D$ to glue together, they would have to agree on ${\rm Pic} (C\cap D)$. In fact, that way one gets a surjective map: $$ {\rm Pic(X)}\to {\rm Pic}(C)\times_{{\rm Pic} (C\cap D)} {\rm Pic}(D). $$ There are obvious restriction maps ${\rm Pic}(X)\to {\rm Pic} (C)$ and ${\rm Pic}(X)\to {\rm Pic} (D)$ and since a line bundle on $X$ is exactly one on each of $C$ and $D$ that agree on the intersection, it follows that ${\rm Pic}(X)$ maps onto the categorical fibered product of ${\rm Pic}(C)$ and ${{\rm Pic} (D)}$ over ${\rm Pic}(C\cap D)$.

EDIT Changed isomorphism to surjective map following Qing Liu's observation. This map is injective if for example the intersection is geometrically connected and geometrically reduced.

share|improve this answer
I think the OP was considering the case in which $C\cap D$ is a single closed point. So, in that case, $Pic(C\cap D)$ is the trivial group, and the resulting $Pic$ is actually the product... –  Qfwfq Mar 2 '11 at 17:57
Is there a proof that avoids using terms like "gluing"? –  shenghao Mar 2 '11 at 18:12
@unknowngoogle: if the intersection is a point then the fibered product is the same as the product. –  Sándor Kovács Mar 2 '11 at 18:14
* product –  Qfwfq Mar 2 '11 at 18:28
@shenghao: I believe that in my proof above you don't ever need to say the word "gluing". There are obvious restriction maps that give you a map from ${\rm Pic}(X)$ to the fibered product. Then you need to think about what it means to have a line bundle on an reducible space. It is not at all gluing in the sense of gluing on open sets. It's a matter of speech. –  Sándor Kovács Mar 2 '11 at 19:09

If the intersection is nice enough, then a line bundle on $X$ consists of an isomporhism class of patching data $(P_C,P_D,\alpha)$ where $\alpha$ is an isomorphism from $P_C|_D$ to $P_D|_C$.

But this does require some assumptions. In the affine case, if $X=Spec(R)$, then you need $R$ to be the pullback of $\Gamma(C)$ and $\Gamma(D)$ over $\Gamma(C\cap D)$.

(I don't actually have an immediate counterexample without this assumption, but certainly the usual proofs require it, and certainly the obvious functor from patching data to line bundles doesn't work without it.)

share|improve this answer

I try a down to earth sketch of proof.

Suppose, w.l.o.g., that $C$ and $D$ are connected. Let

$\pi:W=C\sqcup D \rightarrow X=C \;\;\cup_x\; D$

the map that glues the two components at $x\in X$ (This should be, btw, a coproduct in the category of schemes of $C$ and $D$ along the inclusion of the closed point $x$). Let

$\pi^{\star}:\mathrm{Pic}(X)\rightarrow \mathrm{Pic}(W)=\mathrm{Pic}(C)\times \mathrm{Pic}(D)$

be the induced map.

Now, line bundles $L$ have total spaces $tot(L)$, and two line bundles over a variety are isomorphic iff the corresponding total spaces are isomorphic as "geometric vector bundles".

Remark: if $\widetilde{L}$ is a line bundle on $W$, then $tot(\widetilde{L})$ is the disjoint union of $tot(L')$ and $tot(L'')$ for (uniquely determined) l.b.'s $L'$ and $L''$ on $C$, $D$ respectively.

Remark: given line bundles $\widetilde{L}$ on $W$ and $L$ on $X$, we have $\widetilde{L}=\pi^{\star}(L)$ iff $tot(L)$ is the union (=coproduct), along the fiber over $x$, of the total spaces $tot(L')$ and $tot(L'')$ defined above.


$(C\times\mathbb{A}^1)\sqcup_{x\times\mathbb{A}^1}(D\times\mathbb{A}^1)\cong (C\sqcup_x D)\times\mathbb{A}^1$

then the trivial bundle on $W$ can only come, via $\pi^{\star}$, from the trivial bundle on $X$, i.e. $Ker(\pi^{\star})=0$.

We can see $\pi^{\star}$ is also surjective, because given a bundle $\widetilde{L}$ on $W$, we can glue the corresponding $L'$ and $L''$ along their fibers over $x$, so obtaining a bundle $L$ on $X$ that verifies $\widetilde{L}=\pi^{\star}(L)$.

share|improve this answer

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.