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.

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    $\begingroup$ 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! $\endgroup$
    – Maharana
    Commented Jan 11, 2010 at 23:38
  • $\begingroup$ 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 $Cl(R)$. $\endgroup$
    – Matt
    Commented Apr 12, 2010 at 6:48
  • $\begingroup$ hilbertthm90: The Spec map is not inclusion of open set. $\endgroup$ Commented Apr 12, 2010 at 7:11
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    $\begingroup$ A little off-topic grammar remark. Why say "torson-freeness" and not "torsion-freedom" or something? $\endgroup$ Commented Apr 12, 2010 at 11:55
  • $\begingroup$ 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) $\endgroup$
    – Matt
    Commented Apr 12, 2010 at 22:23

2 Answers 2


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 $\mathbb{Z}/2$.


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 $(-n,-1)$.

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    $\begingroup$ 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. $\endgroup$ Commented Apr 12, 2010 at 13:37
  • $\begingroup$ You are right. Sorry for being so careless. $\endgroup$
    – naf
    Commented Apr 12, 2010 at 14:46
  • $\begingroup$ No problem, thanks for being interested! $\endgroup$ Commented Apr 12, 2010 at 14:58
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    $\begingroup$ 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 $Cl(Y)=\mathbb Z/(2)$. $\endgroup$ Commented Apr 13, 2010 at 17:07
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    $\begingroup$ I have edited the answer. Hope it is clear now. $\endgroup$
    – naf
    Commented Apr 14, 2010 at 14:54

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

  • $\begingroup$ 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. $\endgroup$ Commented Jan 12, 2010 at 19:46

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