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 $t\pi' = s\pi$.

Can one do better than this?

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.

Since p is noetherian, it can be generated by finitely many elements $f_1,\ldots,f_r$. Can one arrange so that $ord_p(f_i) < ord_p(f_{i+1})$?

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).


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.

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 $p$.

  • $\begingroup$ thanks! this was very helpful and clear (but disappointing for me, so I've revised the question...) $\endgroup$ – Yosemite Sam Oct 11 '16 at 0:48
  • $\begingroup$ once again, thanks! In your third remark, I don't see why R/p is a dvr (I must be overlooking something obvious), could you elaborate? $\endgroup$ – Yosemite Sam Oct 11 '16 at 15:07
  • $\begingroup$ I should also say that, in the situation I'm actually in, the divisor corresponding to the ideal p is also Q-Cartier. But I just don't see this condition helping at all with my question. $\endgroup$ – Yosemite Sam Oct 11 '16 at 15:08
  • $\begingroup$ Sorry, that was dumb, $R/p$ can be much worse than $R$. $\endgroup$ – David E Speyer Oct 11 '16 at 17:18
  • $\begingroup$ So, in your actual situation, $R$ is local and $p$ is $\mathbb{Q}$-Cartier? That's very close to principal already, but I would bet you still lose. Let me think. $\endgroup$ – David E Speyer Oct 11 '16 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.