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

- $X$ is
*factorial*if every Weil divisor on $X$ is Cartier. - $X$ 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.

even if its tangent cone at the node is not. – Francesco Polizzi Mar 14 '13 at 14:11destroyfactoriality? So far, I do not know the answer to this question. I also asked this on MO, see mathoverflow.net/questions/115633/… – Francesco Polizzi Mar 14 '13 at 14:18