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](https://mathoverflow.net/questions/98919/factoriality-of-complete-intersections)) or implied ([in this answer to a related question](https://mathoverflow.net/questions/101946/is-there-a-notion-of-local-ample-kahler-cone-for-resolved-singularities/101948#101948)) 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.



Factoriality: local or global?
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.


