# Proving A Discrete Valuation Ring

I am given the definition that $R$, an integral domain, not a field is a DVR if either:

1. R is a local Noetherian ring such that its maximal ideal is principal.
2. There exists an irreducible element $t \in R$ such that every element $r \in R - \{0\}$ can be written uniquely in the form $r=ut^m$ with $u \in R^{\times}$, $m \in \mathbb{N}$.

I am trying to prove that for $p \in \mathbb{Z}$ prime, $\mathbb{Z}_p= \{ \alpha \in \mathbb{Q} | \alpha = \frac{a}{b},$ a and b integers such that p does not divide b $\}$ is a DVR.

In order to understand the DVR better, I would like to prove it using 1 and again using 2.

For 1) I need that R is local Noetherian ring. But to be Noetherian, R has to be finitely generated, and $\mathbb{Q}$ is not. Doesn't this mean that $\mathbb{Z}_p$ is either Noetherian? Then I want to find a maximal ideal that contains all the elements that are not units, right?

For 2) would I want $t$ to be 1 or $a$? And I am rather stuck after this. I think I am fine with uniqueness, but I am not sure how to prove that it works for all.

For 2) would I want t to be 1 or a? if t were 1, every element would be a unit and you would be looking at a field. And $a$ is not a fixed element of the ring, so that is just an invalid choice. Isn't there is a more obvious candidate, an important fellow generating the unique maximal ideal?
• So for 2) we would want t to be p? Then I would want that for $\alpha = \frac{a}{b} = ut^m$. And since the only non-units are of the form $\frac{b}{a}$, I would have \frac{a}{b} *p^m? May 11 '14 at 15:43
• @math1234567 The first sentence is right, but the last sentence isn't clear. Supposing $\frac{a}{b}$ is a unit in the ring, that means that $\frac{b}{a}$ is in the ring too, and for that to happen, $p$ can't divide $a$. Suppose $q$ is some prime besides $p$. Is $q$ a unit in this ring? May 11 '14 at 17:45