4

Is the following statement true?

Let $A$ be an UFD and $p\in A$ prime, then $A_{(p)}$ is a discrete valuation ring.

I think yes: For every element $x$ of $Q(A_{(p)})=Q(A)$, there is a unique $k\in\mathbb{Z}$ such that I can write $x$ as $p^k\cdot\frac{a}{b}$ for some $a,b\in A$ with $p\nmid a,b$. This should give me the well-defined valuation $\nu\colon x\mapsto k$ from $Q(A_{(p)})$ to $\mathbb{Z}$ such that $A$ is the valuation ring associated to $\nu$. Then $\nu$ is surjective, because for every $k\in\mathbb{Z}$ we have $p^k\mapsto k$.

But now I am confused since this would imply that $A_{(p)}$ is Noetherian without further assumptions, which for me seems to come out of nowhere. So is the above argument correct? And if so, is there a more direct way to see that $A_{(p)}$ should always be Noetherian?

Jonathan Gruner
  • 536

2 Answers2

3

Since $p$ is nonzero and prime, $(p)$ is a height one prime, and so $A_{(p)}$ is a one dimensional local UFD, since localizations of UFD's are UFD's. But a one-dimensional UFD is noetherian. Thus, since UFD's are integrally closed, $A_{(p)}$ is an integrally closed noetherian local domain of dimension one, hence is a DVR.

Community
  • 1
Takumi Murayama
  • 10,066
2

Let me complement the nice and abstract existing answer by a concrete one:

Yes, the argument is correct. To see that the ring $A_{(p)}$ is noetherian directly use an argument as you might know it from Euclidean domains.

Let $I$ be an non-zero ideal. Let $a \in I$ be non-zero with minimal valuation, say $k$; then show $I = (a) $, as for $b \in I$ you have $a^{-1}b$ has non-negative valuation and thus is in the ring.

Thus, this is a PID and thus noetherian.

Or, show further that $(a)= (p^k)$ so all non-zero ideals are given by $(p^k)$ with $k$ a natural number and there cannot be an infinite ascending chain as there is no infinite descending chain of natural numbers.

quid
  • 42,835