Let $A$ be a domain satisfying the following condition:

If $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, then we have $$ \mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}. $$

Is $A$ necessarily noetherian?

This question is motivated by these outstanding answers of user26857 and Julian Rosen.

user26857's answer shows that noetherian domains do satisfy the above condition, whereas Julian's answer shows that many non-noetherian domains do not satisfy it.

  • 1
    $\begingroup$ It's true for $1$-dimensional valuation domains. Is it true for valuation domains in general? $\endgroup$ – Badam Baplan Jan 3 at 19:26
  • $\begingroup$ I think there are 1-dimensional valuation domains such that $m^2=m$. $\endgroup$ – user26857 Jan 3 at 23:59
  • $\begingroup$ Yes exactly. A $1$-dimensional valuation domain is Noetherian iff its max ideal is not idempotent iff it satisfies the property in question. $\endgroup$ – Badam Baplan Jan 4 at 0:15

Let's call domains having this property $UPIF$-domains, as in your first post.

Observe that a domain which is locally $UPIF$ is $UPIF$.

Indeed, given $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}$$ we can localize at each of the $\mathfrak{p_i}$ one at a time and use the assumption that $D_{\mathfrak{p_i}}$ is $UPIF$ to deduce that $m_i = n_i$.

From your linked Questions/Answers, we thus have that a locally Noetherian domain is locally UPIF, hence

A locally Noetherian domain is $UPIF$.

However, there are locally Noetherian domains that are not Noetherian.

A classic reference for such examples is section 2 of this paper of Heinzer and Ohm.

For a more recent reference, I'd suggest Loper's article on Almost Dedekind Domains Which Are Not Dedekind. In section 3, he discusses five distinct techniques for producing examples of non-Noetherian domains which are locally discrete valuation rings.


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.