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Once again, the question says it all.

My motivation is the article on factorization I am writing. I want to explain (as well as to understand!) why for normal Noetherian domains of dimension greater than one, the obstruction to factoriality is the nonvanishing of the (Weil) divisor class group $\operatorname{Cl}(R)$, not the Picard group $\operatorname{Pic}(R)$ (equivalently, the Cartier divisor class group).

My understanding is that it is equivalent to find a normal Noetherian domain with vanishing Picard group which is not locally factorial. I would be especially happy to see an example among affine domains, i.e., in which the domain is finitely generated as an algebra over some field.

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Why doesn't something like $\mathbb{C}[x^2,y^2,xy]$ work? – Steve D May 24 '10 at 9:55
To clarify, $\mathbb{C}[x^2,y^2,xy]$ is the fixed subalgebra of $\mathbb{C}[x,y]$ being acted on by $C_2$, where the generator sends $x$ to $-x$ and $y$ to $-y$. This subalgebra is a Noetherian normal domain. Its Picard group is trivial (a theorem of Kang), but its divisor class group is order 2 (a theorem of Nakajima). A reference would be ch. 3 of Benson's "Polynomial Invariants of Finite Groups". – Steve D May 24 '10 at 10:07
Thanks for the clarification. I was thinking of this ring as $\mathbb{C}[x,y,z]/(xy-z^2)$ and got as far as verifying that it is not a UFD (by a theorem of Samuel) and that it is normal. But I don't know anything about Kang's theorem. Anyway, this certainly sounds like the answer: would you please leave it as such and receive your due reward? :) – Pete L. Clark May 24 '10 at 10:21
up vote 16 down vote accepted

If the group $C_2$ acts on $\mathbb{C}[x,y]$ by sending $x$ to $-x$ and $y$ to $-y$, then the fixed subalgebra is the domain $\mathbb{C}[x^2,y^2,xy]$. This is Noetherian and normal. A theorem of Nakajima gives the isomorphism type of the divisor class group (in this case, $C_2$.) A theorem of Kang asserts for such examples (polynomial rings over $\mathbb{C}$, fixed subalgebra of finite group action), the Picard group is always trivial, so one can find many examples.

The reference for all of this is Benson's Polynomial Invariants of Finite Groups. Kang's theorem is Theorem 3.6.1; Nakajima's theorem is Theorem 3.9.2 (and Corollary 3.9.3).


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Thanks again. I have added you to the acknowledgments of my paper. – Pete L. Clark May 24 '10 at 10:43
Thanks for the interesting link, Steve. Even if the full force of Kang's and Nakajima's theorems might not be strictly necessary here, I'm very happy to learn about such powerful and general results . – Georges Elencwajg May 24 '10 at 11:55

Here is a very simple way to find an example: If $R$ is local, then the Picard group is always trivial (because locally free = free here).

So basically you want a local normal domain which is not UFD. Just start with your favorite polynomial ring. Now kill a relation involving elements inside the irrelevant ideal to make sure there are non-unique way to factor: $(xy-z^n)$, $(xy-uv)$, $(xy-uvw)$, etc. and then localize at the irrelevant ideal. (By starting with $xy$ term you make sure that the Jacobian have height at least 2, which together with the fact that $R$ is a hypersurface, guarantees normality)

EDIT: If one wants to use the first kind of polynomials $xy-z^n$ ($n\geq 2$) one needs to make sure that $n$ is prime to the characteristic of the ring. The facts on Picard group and non-UFD are still OK, but to ensure $R$ is normal one needs the Jacobian, which will be $(x,y+ \text{partial derivatives in the second term})$ to be of height $2$.

EDIT2: In certain cases one does not need to localize. Suppose $R=R_0\oplus R_1\oplus\cdots$ is graded and normal, $R_0$ is a field and let $m=R_1\oplus \cdots$. Then by 10.3 of Fossum's book, the maps induced by localization $$ \text{Cl}(R) \to \text{Cl}(R_m)$$ is bijective. The composition: $$\text{Pic}(R) \to \text{Cl}(R) \to \text{Cl}(R_m) $$ is $0$, and the first map is injective, so $\text{Pic}(R)=0$.

In summary, if you pick local or graded normal domains over fields, all you have to to is making sure $R$ is non-UFD.

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Dear Pete, given an arbitrary field $k$, the scheme $X=SpecA$ where $ A=k[X,Y, Z]/(Z^2-XY)$ seems to be an example of what you wish. The divisor class group is of order 2, generated by a ruling of the quadratic cone $X$ (this is proved in Hartshorne's Book-whose-title-need-not-be-spelled-out, page 134).The non factoriality essentially comes from the equation $z^2=xy$, which explicitly tells us that $z^2$ has two factorizations (a short rigorous proof could use a graduation on the relevant ring).See also Hartshorne, page 142 for nullity of Picard group.

Normality of $A$ is a general fact for factorial rings to which a root of a square-free element is added : Matsumura , Commutative Ring Theory, page 65.

This example is the same as Steve D.'s, with another presentation: if you map $\mathbb A^2$ to $\mathbb A^3$ by $x=u^2, y=uv, z=v^2$ you'll see the correspondence between the two points of view. I don't know Kang's nor Nakajima's results-which are cerainly quite interesting- but for the elementary example discussed, they can be bypassed.

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Thanks, Georges. As you can see from my comment above, I was thinking along these lines as well. I am indeed interested in the simplest way to present this very nice example. I'll think more about it and get back to you. – Pete L. Clark May 24 '10 at 10:49
P.S.: Can the characteristic of $k$ be $2$? I get nervous... – Pete L. Clark May 24 '10 at 10:51
Dear Pete, your nervosity is absolutely justified: Hartshorne demands that $char.k\neq2$ and Matsumura that 2 be invertible in the factorial ring. By the way, I'm pleased to learn that you are writing a new article on factorization: do you have a previsional date of uploading, so that I can look for it on your homepage? – Georges Elencwajg May 24 '10 at 11:46
@Georges -- the article in question is an expanded version of I should have a new draft available in the same place within a week. – Pete L. Clark May 24 '10 at 13:09
Great.Thanks, Pete. – Georges Elencwajg May 24 '10 at 14:02

This is probably too long for a comment, but it ties in with the answers already given.

In dimension 2, factorial local rings are very rare. There is a fascinating discussion, with references, in Hartshorne, V.5.8. In particular, over an algebraically closed field $k$ of characteristic not equal to 2,3,5, the ring $k[[x,y,z]]/(x^2+y^3+z^5)$ is the only 2-dimensional factorial non-regular normal complete local ring. (I remember being spellbound by the statement when I read Harstshorne for the first time and memorizing it as a perfect example of impressive and impenetrable mathspeak). This ring is the completed coordinate ring of the Klein – du Val singularity $\mathbb{A}^2/\Gamma$ of type $E_8$, where $\Gamma$ is the binary icosahedral group. Among the $A, D, E$ root systems, $E_8$ is the only type for which the determinant of the Cartan matrix is 1 (equivalently, the weight and root lattices coincide). For all other Kleinian singularities, you can see very explicitly that the unique factorization fails (see Shafarevich, IV.4.3 of 2nd edition or Springer, Invariant Theory):

$A_n: x^2+y^2+z^{n+1}=0, \quad (x+iy)(x-iy)=-z^{n+1}$

$D_n: x^2+yz^2+z^{n-1}=0, \;n\geq 4 \quad x^2=-z^2(y+z^{n-3})$

$E_6: x^2+y^3+z^4=0 \quad (x+iz^2)(z-iz^2)=-y^3$

$E_7: x^2+y^3+yz^3=0 \quad x^2=-y(y^2+z^3)$

$E_8: x^2+y^3+z^5=0 \quad$ factorial

As has already been mentioned, all these rings are normal domains with trivial Picard group, since they are the $\Gamma$-invariant subrings for a finite group $\Gamma.$

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I think that in the equation for $D_n$ the term $yz^2$ might have to be $yz$, otherwise it seems that $x/z$ would be integral, contradicting normality. – M P Sep 17 '11 at 20:45

These results hold for any field and for many semigroup rings. The first reference is a paper of mine from 1978 or so in the Canada. Math. J (I think). David F. Anderson

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Pete: I recall answering the question: Must an almost factorial domain be locally factorial? You can find the answer at: and some simpler examples at The examples are all Noetherian, normal and locally non-factorial. Now for a local domain the Picard group is zero. Muhammad

PS. I checked out It looks different from an earlier version that I once saw.

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@Muhammad: Thanks for this. The latest version of the article is As the name suggests, nothing has been done with it in the last couple of years. – Pete L. Clark Jul 23 '12 at 15:15
Thanks Pete, that seems to be what I saw. Muhammad – Muhammad Jul 25 '12 at 17:18

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