# A Non-Noetherian Subring of a Polynomial Ring

Posted by Jason Polak on 28. January 2015 · 1 comment · Categories: math

Sometimes there are wild beasts hiding in the tamest of rings. Let's flush one out! Let $k$ be a field. Consider the following subring $R$ of $k[x,y]$ consisting of polynomials all of whose monomials are never of the form $x^i$ for $i\geq 1$. So $f = xy + y^2 + x^2y\in R$ but $f = xy + x^2$ is not. True or false: $R/(xy) \cong K[y]$?

If you thought the answer is 'true', think again! We cannot just "set $xy= 0$" — in fact, $x^2y\not\in (xy)$, since $x\not\in R$. This curious example of Graham Evans provides all sorts of interesting phenomenon, and is a good example to keep in mind. Do you think you understand what makes Krull's principal ideal theorem tick? Let $M$ be the prime ideal consisting of all polynomials with constant term zero. It is in fact maximal, and is minimal over the ideal $(y)$. Yet, ${\rm rank}(M) \geq 2$! If $R$ were Noetherian, Krull's principal ideal theorem would state ${\rm rank}(M)\leq 1$. So, obviously $R$ is not Noetherian. Can you find an infinitely generated ideal? Yet, it's easy to see that $R$ still satisfies the ascending chain condition on principal ideals, so the ascending chain condition on principal ideals is insufficient for the principal ideal theorem to hold.

### 1 Comment

1. Hi,
I guess the easiest example of non-noetherian subring of a polynomial ring (in one variable only!) is the classical ring of integer-valued polynomials:
Int(Z)={f\in Q[X] | f(Z)\subset Z}
strictly contained in between Z[X] and Q[X]. It can be easily shown that the following ideal is not finitely generated:
{f\in\Int(Z) | f(0)\in 2\Z}
More precisely, Int(Z) is a 2 dimensional Prufer non-noetherian domain. See the book of Cahen and Chabert about the topic.