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

Let $R$ be a local normal domain, and let $P \in Spec (R)$. It is well known that $Cl(R) \to Cl(R_P)$ is surjective. However, I do not know any example where $Cl(R)$ is torsion-free, but $Cl(R_P)$ is not (we consider $(0)$ to be torsion-free). So:

If the class group of $R$ is torsion-free, must all the class groups of local rings of $R$ torsion-free? What if $R$ is not local?

I do not have a motivation for this, but it just seems an intriguing question.

share|improve this question
1  
For the longest of time I couldn't even find an example of $R$ other than the affine quadric threefold such that it has torsion-free Class group. But now that I have some of them, they all turn out with local class groups $0$. Your question is so tough! –  Maharana Jan 11 '10 at 23:38
    
I'm not incredibly comfortable with this stuff, so forgive me if this is way off. $R\hookrightarrow R_P$ induces $Spec(R_P)\to Spec(R)$ and then $Cl(R)\to Cl(R_P)$ is the pullback under the previous map. But geometrically isn't the Spec map an inclusion of an open set, so the pullback is just taking a divisor on the open set and considering it as a divisor on the whole space. There is probably some subtlety I'm missing, but it seems if the above is correct that since it is really the "same" divisor, that $pD=0$ on the open should imply $pD=0$ on $Cl(R)$. –  Matt Apr 12 '10 at 6:48
    
hilbertthm90: The Spec map is not inclusion of open set. –  Hailong Dao Apr 12 '10 at 7:11
1  
A little off-topic grammar remark. Why say "torson-freeness" and not "torsion-freedom" or something? –  Gerald Edgar Apr 12 '10 at 11:55
    
Sorry. That was sloppy wording due to lack of space. It is a isomorphism onto an open subset. This group is an isomorphism invariant, so is it possible to then just work on this open set? I realized the subtlety from the last comment. If $pD \sim (f)$ on U (the open set), then we'd hope $pD\sim (f)$ considering $f\in K(SpecR)=K(U)$. It seems possible that on the whole space it could have more vanishing and poles. Is there an example of that? It might lead to your example. If there isn't it might lead to a proof (I suspect the latter) –  Matt Apr 12 '10 at 22:23
show 2 more comments

2 Answers

up vote 1 down vote accepted

Perhaps something like the following works (I have not checked all the details):

Let $C$ be a smooth plane conic and let $Y$ be the projective cone over $C$. Then $Cl(Y) = \mathbb{Z}$ but the class group of the local ring of the vertex of $Y$ is $\mathbb{Z}/2$. Let $X$ be affine cone over $Y$. Then the class group of the local ring $R$ of the vertex of $X$ is $\mathbb{Z}$ but it seems that the class group of $R$ localised at the prime ideal corresponding to the cone over the vertex of $Y$ is $\mathbb{Z}/2$.

EDIT

The above is wrong as pointed out by Hailong Dao in his comment. I try to fix it below:

Let Y be as above i.e. the singular quadric in $\mathbb{P}^3$ given by the equation $x^2 + y^2 +z^2 = 0$. It may be viewed as the toric surface given by the complete fan with rays passing through $(1,0)$, $(0,1)$ and $(-1,-2)$. Then $Cl(Y) = \mathbb{Z}$ and $Pic(Y)$ is of index $2$ in $Cl(Y)$. Let $Y'$ be the blowup of $Y$ at a non-singular torus fixed point. We may view Y as the surface obtained from the fan for $Y$ by adding the ray through the point $(1,1)$. Let $\pi:Y' \to Y$ be the blowup map and let $E$ be the exceptional divisor. Then $Cl(Y') \cong \mathbb{Z} \oplus \mathbb{Z}$ and $Cl(Y')/\mathbb{Z}E = Cl(Y)$.

Let $H$ be an ample divisor on $Y$. Then for $n >> 0$, $H':= n\pi^*(H) - E$ is an ample divisor on $Y'$ (this is true for the blowup of a point on any surface). Note that $Cl(Y')/\mathbb{Z}H' \cong \mathbb{Z}$, so it is torsion free. SInce $Y'$ is a projective toric surface and $H'$ is an ample divisor, it follows that $H'$ is very ample and gives a projectively normal embedding of $Y'$ in $\mathbb{P}(H^0(Y',\mathcal{O}(H')))$.

As before, we now let $X$ be the cone over $Y'$ and let $R$ be the local ring of the vertex. We let $P$ be the prime ideal corresponding to the cone over the singular point of $Y'$.

($Cl(Y)$ and $Cl(Y')$ can be computed by hand or using the toric description I gave and the results in Fulton, Toric Varieties, Sections 3.3, 3.4; the fact that an ample divisor on a projective toric surface is very ample is an Exercise at the bottom of p.70.)

Note that by letting $R$ be the coordinate ring of $Y' \backslash D$ , where $D$ is a general divisor linearly equivalent to $H'$ (so not containing the singular point) one gets a normal 2-dimensional (non-local) ring with $Cl(R)= \mathbb{Z}$ and with a prime ideal $P$ such that $Cl(R_P)=\mathbb{Z}/2$ . In all of the above one can replace $2$ by any integer $n>1$ (by considering the projective cone over the rational normal curve on degree $n$, or, in the toric description, replacing $(-2,-1)$ by $(-n,-1)$.

share|improve this answer
    
The class group of the local ring the origin of the cone over Y is $Cl(Y)/\mathbb Z(H)$, where $H$ is the hyperplane section. So I don't think your example works. –  Hailong Dao Apr 12 '10 at 13:37
    
You are right. Sorry for being so careless. –  ulrich Apr 12 '10 at 14:46
    
No problem, thanks for being interested! –  Hailong Dao Apr 12 '10 at 14:58
    
Hi again, thanks for the new post. This looks really interesting, but I am a little confused about your $Y$. Do you mean something like $Y=Spec k[x,y,z]/(x^2+y^2+z^2)$? Then I think $Cl(Y)=\mathbb Z/(2)$. –  Hailong Dao Apr 13 '10 at 17:07
    
I have edited the answer. Hope it is clear now. –  ulrich Apr 14 '10 at 14:54
show 2 more comments

The two papers cited below might be of some interest to you:

Theorem 3.3 of (1) says that for a Krull domain $R$, $Cl(R)/Pic(R)$ is torsion iff local class groups are torsion at the maximal ideals. It is also a remark before Theorem 13 of (2), attributed to Chouinard, that any abelian group appears as the Class group of some local Krull domain. So for constructing an example as you want we may choose a local Krull domain $R$ with $Cl(R)$ a free abelian group and try to see if we can manage the image of $Pic$ in $Cl$ to be of finite index. I am unable to construct such an example right away, however its existence seems morally possible to me.

(1) Anderson, Globalization of some local properties.., MR0652428

(2) Bouvier, The local class group.. , MR0681946

share|improve this answer
    
Thanks for the reference! If R is local then Pic(R)=0, so (1) does not help. I will try to think more about applying Theorem 3.3. –  Hailong Dao Jan 12 '10 at 19:46
add comment

Your Answer

 
discard

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.