What is an easy example of non-Noetherian domain?

Question

Keep in mind, I'm strictly an amateur, though a very old one. I learned about imaginary numbers barely two years ago and ideals a year ago, and I'm still decidedly a novice in both topics.

In the university library, I was looking at Modules over Non-Noetherian Domains by Fuchs and Salce and I couldn't really understand anything. I'm also looking at the "Questions that may already have your answer," but if they do, it's not in a way that I can understand.

Then I thought what about a finite ring, like maybe $\mathbb{Z}_{10}$, but that's only created more questions, like: can a finite ring be non-Noetherian? Although $5 = 5^n$ for any $n \in \mathbb{Z}_{10}$ besides $0$, we're still dealing with only the ideal $\langle 5 \rangle$, right? There's no ascending chain of ideals even though some numbers in this domain have infinitely many factorizations, right? It is a Noetherian ring after all, right?

My question, it seems, has then become if it's possible for a non-Noetherian ring to be within the comprehension of a dilettante such as myself, or must it necessarily be esoteric and exotic?

Answer 1

Being "Noetherian" can be read as a ring for which any ascending chain of ideals has a "biggest ideal", one that contains all the others but is only contained by ideals which are equal to itself.
To be non-Noetherian, the ring simply needs to have an infinite ascending chain of ideals. The ring of algebraic integers, for example, has the infinite chain of ideals generated by $2^{1/{2^{n}}}$.
That is, $$\langle \sqrt{2} \rangle \subset \langle \sqrt[4]{2} \rangle \subset \langle \sqrt[8]{2} \rangle \subset \langle \sqrt[16]{2} \rangle \subset \dots$$ forms a chain without a "biggest link".

Answer 2

For a field $k$ the ring $k[x_1,x_2,\dots]$ of polynomials with infinite indeterminates is non-Noetherian because you can take the ascending chain

$$(x_1)\subset (x_1,x_2)\subset(x_1,x_2,x_3)\subset\cdots$$

And also to answer one of your questions, a finite ring must be Noetherian because an equivalent definition of Noetherian is "every ideal is finitely generated", so if $R$ is finite then each ideal $I$ is finite and in particular it's generated by itself, so every ideal is finitely generated.

Answer 3

The ring of integer-valued polynomials (the subring of $\mathbf Q[x]$ of polynomials which take integer values at integers) is another example of a non-noetherian integral domain.

The ring of continuous functions on $[a, b]$ is yet another example (it's not an integral domain).

Answer 4

You should check out Richard Borcherds' very accessible (at least to begin with) and entertaining online lecture series on Commutative Algebra. In his lecture on Noetherian Rings he gives several very instructive examples of rings which are all somewhat similar, and shows that some are Noetherian while others aren't.

Here are his examples:

• polynomials over the reals (Noetherian)
• analytic functions over the reals (Non-Noetherian)
• analytic funtions on [-1,1] (Noetherian)
• analytic functions on (-1,1) (Non-Noetherian)
• functions analytic in a neighbourhood of zero (Noetherian)
• functions that are smooth in a neighbourhood of zero, i.e. germs of smooth functions (Non-Noetherian)
• formal power series over the reals (Noetherian)