11
$\begingroup$

Let R be a local, normal domain of dimension 2. Suppose that R contains a finite field. I am interested in knowing when the class group of R is torsion. In characteristic 0, this is known to be related to R being a rational singularity.

Lipman showed that if X is a desingularization of Spec(R), then one has an exact sequence:

$0 \to Pic^{0}(X) \to Cl(R) \to H $

Here $Pic^{0}(X)$ is the numerically trivial part of the Picard group of $X$, and $H$ is a finite group. Thus the second one is torsion if and only if the first one is. I do not have much understanding of the first group, unfortunately.

Does anyone know an answer or reference to this? Does anyone know an example in positive characteristic such that $Cl(R)$ is not torsion? Thanks a lot.

$\endgroup$
2
  • $\begingroup$ I was going to point you to this nice paper I read recently about Frobenius and torsion in the class group ... then I saw who the poster was! $\endgroup$ Nov 29, 2009 at 23:20
  • $\begingroup$ Hi Graham! Yes, unfortunately we need dimension 3! I think a general answer is probably not easy, but there are many smart people here. $\endgroup$ Nov 30, 2009 at 2:07

2 Answers 2

26
$\begingroup$

As requested in the comments, here's an example of a local, normal $2$-dimensional domain R in positive characteristic such that $\mathrm{Cl}(R)$ is not torsion: choose an elliptic curve $E \subset \mathbf{P}^2$ over a field $k$ such that $E(k)$ is not torsion, and take R to be the local ring at the origin of the affine cone on $E$ (i.e., $R = k[x,y,z]/(f)_{(x,y,z)}$ where $f$ is a homoegenous cubic defining $E$). This can be done over $k = \overline{\mathbf{F}_p(t)}$.

Proof: The normality follows from the fact that R is a hypersurface singularity (hence even Gorenstein) and isolated and $2$-dimensional (hence regular in codim 1). Blowing up at the origin defines a map $f:X \to \mathrm{Spec}(R)$. One can then show the following: $X$ is smooth, and $X$ can be identified with the Zariski localisation along the zero section of the total space of the line bundle $L = \mathcal{O}_{\mathbf{P^2}}(-1)|_E$ (these are general facts about cones). By Lipman's theorem, it suffices to show that $\mathrm{Pic}^0(X)$ contains non-torsion elements. As $X$ is fibered over $E$ with a section, the pullback $\mathrm{Pic}^0(E) \to \mathrm{Pic}^0(X)$ is a direct summand. As $\mathrm{Pic}^0(E) \simeq E(k)$ has non-torsion elements by assumption, so does $\mathrm{Pic}^0(X)$.

Also, an additional comment: In general, Lipman's theorem tells you that $\mathrm{Cl}(R)$ is torsion if and only if $\mathrm{Pic}^0(X)$ is torsion. Now $\mathrm{Pic}(X) \simeq \lim_n \mathrm{Pic}(X_n)$ where $X_n$ is the $n$-th order thickening of the exceptional fibre $E$. Because we are blowing up a point, the sheaf of ideals $I$ defining $E$ is ample on $E$. The kernel and cokernel of $\mathrm{Pic}(X_n) \to \mathrm{Pic}(X_{n-1})$ are identified with $H^1(E,I|_E^{\otimes n+1})$ and $H^2(E,I|_E^{\otimes n+1})$. As $I|_E$ is ample, it follows that the system "$\lim_n \mathrm{Pic}(X_n)$" is eventually stable. Thus, $\mathrm{Pic}(X) \simeq \mathrm{Pic}(X_n)$ for $n$ sufficiently big. As $X_n$ is a proper variety, it follows that if we are working over a finite field (resp. an algebraic closure of a finite field), then $\mathrm{Pic}^0(X)$ is finite (resp. ind-finite).

$\endgroup$
7
  • $\begingroup$ Excellent, thanks! Is there a reference about your fact on elliptic curves over big fields? $\endgroup$ Jan 17, 2010 at 5:07
  • $\begingroup$ Also, why $X_n$ proper implies $Pic^0$ finite? $\endgroup$ Jan 17, 2010 at 5:13
  • $\begingroup$ Take any old elliptic curve E over F_q, and view it as an elliptic curve E' = E \times_{F_q} k over its function field k = k(E). The generic point in E'(k) will be non-torsion (as all the torsion is defined over F_q-bar). $\endgroup$
    – Bhargav
    Jan 17, 2010 at 11:39
  • $\begingroup$ The connected component $\mathrm{Pic}^0(Y)$ of the Picard group $\mathrm{Pic}(Y)$ is representable by a group scheme of finite type when Y is a proper variety, so its F_q-points form a finite set. (For spaces like the X_n appearing above, the finiteness follows from that of $\mathrm{Pic}^0(X_0)$ using the description of the kernel and cokernel I mentioned in the answer). $\endgroup$
    – Bhargav
    Jan 17, 2010 at 11:43
  • $\begingroup$ Beautiful! I wish I can upvote more. Please upvote this answer, this is really what MO is about. $\endgroup$ Jan 17, 2010 at 15:52
2
$\begingroup$

I recently found some references: Theorem 4.5 of this paper and Theorem 4 + next Corollary of this paper which says:

If $(R,m,k)$ is a complete normal local domain of dimension $2$ such that $k$ is the algebraic closure of some finite field, then $Cl(R)$ is torsion!

It remains open what happens in other situations.

$\endgroup$
5
  • $\begingroup$ You mean only the case $k=\mathbb F_q$ remains? This case follows from the case $\overline{F}_p$ at once, I think. But what if $k$ is algebraically closed and not equal to $\mathbb F_p$? $\endgroup$
    – VA.
    Jan 17, 2010 at 0:02
  • $\begingroup$ Why would the $F_p$ follow? The class group may get smaller if one enlarge the field. But my comment was not very clear, thanks for pointing it out $\endgroup$ Jan 17, 2010 at 0:29
  • $\begingroup$ The statement will not be true for big fields k. Take the (completed) affine cone Y on an elliptic curve E in P^2 over k. Blowing up at the origin resolution X -> Y. On the other hand, X is the total space of O(-1) over E (suitably completed). In particular, Pic^0(X) contains Pic^0(E) =~ E(k) as a subgroup, and this can certainly have non-torsion elements if k is not algebraic over the prime field. $\endgroup$
    – Bhargav
    Jan 17, 2010 at 1:38
  • $\begingroup$ @bhargav: would you write up your answer? I want to upvote it. $\endgroup$ Jan 17, 2010 at 2:03
  • $\begingroup$ The links to sciencedirect.com and springerlink.com both appear to be broken. Perhaps you could take a look, whenever possible... $\endgroup$ May 5, 2022 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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