Let R be a noetherian normal domain (if it makes any difference, I'm happy to assume R is also local).


If $p$ is a height one prime, then the localization $R\_p$ is a dvr, hence the maximal ideal $pR\_p$ is principal, generated by a uniformizer $\pi$.


It's easy to see we can pick $\pi \in p$. Similarly, if $\pi' \in p$ is another uniformizer then there exist $s,t \in R - p$ such that $t\pi' = s\pi$.



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> Can one do better than this?
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More precisely, is there a uniformizer $\pi\_0 \in p$ such that any other uniformizer $\pi'$ is of the form $\pi' = s\pi\_0$ for $s \in R - p$?


EDIT: David's answer shows that an affirmative answer to the question above actually implies p itself is principal (which is very strong). So here's a weaker question.



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> Since p is noetherian, it can be generated by finitely many elements $f\_1,\ldots,f\_r$. Can one arrange so that $ord\_p(f\_i) < ord\_p(f\_{i+1})$?
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The motivation for this question comes from the following standard example. Take $R = k[x,y,z]/(z^2 - xy)$. Take $p = (x,z)$. This is a height one prime ideal. Here $z$ is a uniformizer. Although $p$ is not principal, one needs only one element of order one to generate it.


EDIT 2: I would also happily assume that R is moreover complete and an algebra over a field (of say characteristic zero).



uniqueness of uniformizers
This is the same as asking that $p$ is principal. In one direction, if $p = (\pi)$, then take $\pi$ to be the uniformizer. 


In the reverse direction, suppose $\pi$ has the stated property. I claim that $p = (\pi)$. Let $f$ be a nonzero element of $p$; we must show that $f$ is a multiple of $\pi$. Since $R$ is a noetherian domain, $\bigcap p^n = (0)$, so there is some $n$ for which $f \in p^n \setminus p^{n+1}$. Thus, $f = \pi\_1 \pi\_2 \cdots \pi\_n$ where each $\pi\_i \in p \setminus p^2$, and is thus a uniformizer. By the hypothesis, each $\pi\_i$ is of the form $s\_i \pi$, so $f = (\prod s\_i) \pi^n$. We have shown that $\pi$ divides $f$.


A Krull domain has all height one primes principal iff it is a UFD. A normal noetherian domain is Krull. Thus, your condition holds for all height one primes if and only if $R$ is a UFD.




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Response to the new question: Let $p\_k = \{ x \in R : \mathrm{ord}\_p(x) \geq k \}$. So $p = p\_1$, but note that $p\_2$ is probably larger than $p^2$. (In your example, $y \in p\_2$ but $y \not \in p^2$.) I claim your new condition is equivalent to saying that $p/p\_2$ is a free $R/p$ module. Proof: If $p = \langle f\_1, f\_2, \ldots, f\_r \rangle$ as you propose, then $p/p\_2 = \langle f\_1 \rangle$. Conversely, if $p/p\_2 = (R/p) f$, then lift $f$ to $f\_1 \in p$ and $p/(R f\_1 + p\_2)=0$. Choosing generators for $p\_2$, we have your claim.


I don't know a general result about when $p/p\_2$ is free, but I see three observations: (1) It is not always true. Take $R = k[w,x,y,z]/(wz-xy)$ and $p = \langle w,x \rangle$, so $R/p = k[y,z]$. Then $p\_2 = \langle w^2, wx, x^2 \rangle$ and $p/p\_2$ is the $k[y,z]$ module on generators $w$ and $x$ subject to the relation $z \cdot w = y \cdot x$, which is not true. (2) It is true in Dedekind domains, since $p\_2 = p^2$ and $p/p^2$ is a field. (3) It is true if $R$ is normal and local of dimension $2$, since then $R/p$ is a dvr. This argument was false. I'll modify it to (3') The statement is true if $R$ is normal and $R/p$ is a dvr. Over a dvr, every torsion free finitely generated module is free. The $R/p$ module $p/p\_2$ is torsion free, and is rank $1$ since $R$ is regular at $p$.


