I don’t feel like an all out post, but I did need to use this fact today and it relates to what I’ve been doing, so may as well post the proof. Let be an integrally closed Noetherian domain. Then .
We start with a Lemma: Every prime divisor of a non-zero principal ideal has height 1.
Suppose is a prime divisor of . Then there is an element such that . Define . So . Thus and .
If we had then the standard determinant trick (see a proof a Nakayama’s Lemma) would give us that is integral over contradicting being integrally closed. So and hence and so is a DVR which means dimension of is 1 which means . This gives the lemma.
We will prove that for with we have for all height 1 primes means that is actually in .
Let be the prime divisors of and let be a primary decomposition of where is -primary. But each has height 1 by the Lemma above, so . Thus .
Maybe a short discussion could be useful at this point. The use of this fact came up when trying to prove something about a normal point. This says that at a normal point on a variety, if you want to check if a function regular, you only need to check that it is regular on all codimension 1 subvarieties through that point. This almost immediately gives that if you have a regular function on () where is a normal point. Then the function actually extends to be regular on all of . A function on a dim variety cannot blow up at just a single normal point.