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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.

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But to be Noetherian, R has to be finitely generated, and Q is not That depends on what module we're referring to. Q is certainly not finitely generated by Z, but it is finitely generated by Q.

The bigger issue is that we are not talking about either of these: we are talking about a subring R of Q. If it is to be Noetherian, its ideals will all be finitely generated as modules over R. You are describing the localization of Z at the (complement of) the prime ideal (p). Localizations of commutative Noetherian rings are also commutative. This can be proven by using facts relating the ideals of the two rings.

Then I want to find a maximal ideal that contains all the elements that are not units, right? Sure, and that ideal is obvious, right? Which fractions can't you "flip over" to get their inverses?

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?

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  • $\begingroup$ 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? $\endgroup$ – math1234567 May 11 '14 at 15:43
  • $\begingroup$ @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? $\endgroup$ – rschwieb May 11 '14 at 17:45

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