# Divisible By Arbitrary Powers? Not Noetherian

I've spent countless hours thinking about associative rings. Yet, during my research today I read a paper by C.U. Jensen [1] and came across this elementary fact that I never thought about: if $R$ is an integral domain and $x,y$ are nonunits with $y$ divisible by every positive power of $x$ then $R$ is not Noetherian. Perhaps I've used this unconsciously, but I had to take a second to prove it.

Here's the proof: if $y = z_1x = z_2x^2 = \cdots$ then the ideal $(z_1,z_2,\cdots)$ cannot be finitely generated, because $z_n = r_1z_1 + \cdots + r_{n-1}z_{n-1}$ and $y = z_ix^i$ for all $i$ implies that $x$ is actually invertible.

This fact is curious, because it plays into constructing some nonstandard models of the integers as I learned from Jensen's paper: If $\Ucl$ is a nonprincipal ultrafilter on the natural numbers $\N$, then the product $\Z^\N/\Ucl$ is one such nonstandard model: it satisfies precisely the same first-order sentences that $\Z$ does. Yet, it's not Noetherian, because the element represented by $(2,4,8,16,32,64,…)$ is not a unit and divisible by every power of $2$.

In fact, not only is the ultraproduct $\Z^\N/\Ucl$ not Noetherian, it actually has global dimension two. So, neither Noetherian nor having global dimension one is expressible in first-order logic.

It's amazing what basic facts you can still come across in this era. Perhaps someone should write the book 100 Quick Facts You Didn't Know About Rings?!

[1] Jensen. Peano rings of arbitrary global dimension. Journal of the London Mathematical Society, 1980, 2, 39-44