Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be an algebraic variety. I have read the following definitions:

  1. $X$ is factorial if every Weil divisor on $X$ is Cartier.
  2. $X$ is locally factorial if all its local rings are unique factorisation domains.

I am comfortable with each of these. However, I have also seen it stated (in this question) or implied (in this answer to a related question) that these are equivalent, and I'm not sure about this.

For example, consider the local model for a threefold node, $\mathrm{Spec}\left( \mathbb{C}[x,y,z,w]/(xy - wz) \right)$. Its local ring at the origin is not a UFD, as $xy = wz$. This corresponds to the existence of non-Cartier divisors like $\{x = z = 0\}$.

But this singularity occurs in varieties which do not have non-Cartier divisors. Consider, for example, the quintic threefold given by projectivising the equation ($xy - wz +$ generic terms up to order five) $=0$. I know for a fact that this does not have non-Cartier divisors, yet it has a node.

So, are 1. and 2. really equivalent? If so, the higher-order terms in the above example must restore factoriality of the local ring, right? I'd been under the impression that such higher-order terms don't change the local structure; this might be where I've gone wrong.

share|cite|improve this question
The higher order terms can indeed affect the factoriality of the ring. In your example, the completion of the local ring, or even the analytic local ring, is not a UFD but the local ring itself is. – ulrich Mar 14 '13 at 12:19
In fact, a threefold of degree $d$ with a single node and no other singularities is factorial if and only if $d\geq 3$, see also this question… In other words, if $d \geq 3$ then $X$ is factorial, even if its tangent cone at the node is not. – Francesco Polizzi Mar 14 '13 at 14:11
By the way, I'm also interested in the other way around. Assume that $X$ has only a isolated singularity such that the corresponding tangent cone is factorial. Then, is $X$ factorial? In other words, we know by the examples above that higher-order terms can restore factoriality of the local ring. But can they also destroy factoriality? So far, I do not know the answer to this question. I also asked this on MO, see… – Francesco Polizzi Mar 14 '13 at 14:18
"...analytic local ring, is not a UFD but the local ring itself is." This must be closely related to the fact that small resolutions always exist in the analytic category, but not in the algebraic category? Francesco, thank-you for your useful comments and the links. – Rhys Davies Mar 14 '13 at 14:35

1 Answer 1

up vote 11 down vote accepted

It's a standard result in commutative algebra that every noetherian integral domain is a UFD if and only if every prime ideal of height 1 is principal. When applied to the local rings of X this gives exactly the equivalence above.

share|cite|improve this answer
Thank-you Angelo, that's very concise. – Rhys Davies Mar 14 '13 at 14:59

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.