Why is UFD a Krull domain?

Matsumura mentions this as if it is obvious, and I can't find this result anywhere. Am I missing something obvious here?

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Just found a proof in Kaplansky's Commutative Rings. I can't seem to delete this question, though. – ashpool Feb 18 '11 at 16:05
I don't think it's necessary or even desirable to delete the question. It's a very fair question for a reader of Matsumura (a standard text) to have. The fact that it's answered somewhere else (it had better be, and of course it is) is nothing against answering it here. – Pete L. Clark Feb 18 '11 at 16:08

Just for completeness: A Krull domain is an integral domain $D$ with field of fractions $K$ for which there is a family $\mathcal{F}=\{R_{\lambda}\}_{\lambda\in\Lambda}$ of discrete valuation rings of $K$ such that:

1. $D = \mathop{\cap}\limits_{\lambda\in\Lambda}R_{\lambda}$; and
2. For every $x\in K$, $x\neq 0$, there are at most a finite number of $\lambda\in\Lambda$ such that $v_{\lambda}(x)\neq 0$.

So, assume that $D$ is a UFD. For each (class of associated) irreducible element $\pi$ of $D$, localizing away from $\pi$ (only allowable denominators are prime to $\pi$) gives you a DVR with valuation given by $(\pi)$. The intersection of all of these DVRs, viewed as subrings of the field of fractions of $D$, is exactly $D$ (because the only elements in the field of fractions that can be written with denominators prime to all irreducibles are the elements of $D$), and every element of the field of fractions has nonzero $\pi$-valuation only at finitely many $\pi$ (write the fraction in reduced terms: only those $\pi$ that show up in the numerator or the denominator give you nonzero valuation). So $D$ is a Krull ring.

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