2
$\begingroup$

I asked this question on Mathematics Stack Exchange but got no answer.

Here is the question:

Let $A$ be a domain (that is, a commutative ring with one in which the condition $ab=0$ implies $a=0$ or $b=0$).

Assume that $A$ has the following property:

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 $$ \mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}. $$

Does it follow that $A$ is locally noetherian?

Here are a couple of comments:

In this answer Badam Baplan pointed out that locally noetherian domains do have the indicated property, and that that some non-noetherian domains are locally noetherian (the first example seems to have been given by N. Nakano in 1953).

Previously user26857 had observed that noetherian domains have this property, and Julian Rosen had shown that many non-noetherian domains do not have the property.

$\endgroup$
  • 1
    $\begingroup$ The converse does hold for the class of Prüfer domains, i.e. a Prüfer domain is locally noetherian iff it satisfies the property in question. $\endgroup$ – Badam Baplan Jan 17 at 16:34
5
$\begingroup$

No, this property does not imply that $A$ is locally Noetherian.

For example, let $F\subset L$ be an extension of fields, and let $A=F+XL[[X]]$ (that is, $A$ is the set of power series over $L$ whose constant term belongs to $F$). Then, $A$ is a one-dimensional local domain with maximal ideal $\mathfrak{m}=XL[[X]]$. Hence, any product of nonzero prime ideals is equal to $\mathfrak{m}^n$ for some integer $n$, and these are all distinct since $\mathfrak{m}^n=X^nL[[X]]$, so your property holds.

However, $A$ is not necessarily Noetherian: indeed, if $\mathfrak{m}$ is finitely generated (as an ideal over $A$) then $L$ will have a finite generating set as an $F$-vector space. In particular, if $[L:F]=\infty$ then $A$ is not Noetherian (and so, being local, it is neither locally Noetherian).

$\endgroup$
  • $\begingroup$ Thank you very much for this wonderful answer! (Minor typo: "if $\mathfrak{m}$ IF finitely generated" should be "if $\mathfrak{m}$ IS finitely generated".) $\endgroup$ – Pierre-Yves Gaillard Jan 17 at 11:52
  • $\begingroup$ You're welcome! (I've fixed the typo.) $\endgroup$ – Dario Spirito Jan 17 at 18:32

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.