1. ## Non noetherian rings

What's a really simple example of a non-noetherian ring?

Would the ring B of all algebraic integers be such an example?

2. Try $\displaystyle {\mathbf Z}[1/2]$: polynomials in 1/2, or if you prefer, rational numbers whose denominator is a power of 2. Consider the ascending chain of ideals $\displaystyle \left\langle 1/2 \right\rangle \subset \left\langle (1/2)^2 \right\rangle \subset \left\langle (1/2)^3 \right\rangle \subset \cdots$.

3. ## A not-so-simple non-Noetherian ring

I found the following example in a Wikipedia article:

"The ring of polynomials in infinitely-many variables, X1, X2, X3, etc. The sequence of ideals (X1), (X1,X2), (X1,X2, X3), etc. is ascending, and does not terminate."

The base ring is not mentioned, but I think the example holds for polynomials over any commutative ring (take the real numbers I suppose for concreteness).

The article also mentions that "non-Noetherian rings tend to be very 'large' ", whatever large means. I suppose that might mean, don't expect to find very many mundane, intuitively accessible examples of non-Noetherian rings.

As for your algebraic integers, sorry, I'm no help there.

http://en.wikipedia.org/wiki/Noetherian_ring

4. Originally Posted by rgep
Try $\displaystyle {\mathbf Z}[1/2]$: polynomials in 1/2, or if you prefer, rational numbers whose denominator is a power of 2. Consider the ascending chain of ideals $\displaystyle \left\langle 1/2 \right\rangle \subset \left\langle (1/2)^2 \right\rangle \subset \left\langle (1/2)^3 \right\rangle \subset \cdots$.
Aren't all your ideals equal here? Since 1/2 is in the base ring, 1/4 is in your first ideal, and since 2 is in the base ring, 1/2 is in your second ideal, making them equal, and a similar logic can be applied to the others. In fact, since all of these ideals contain 1, they are all equal to the entire ring, right? Or am I confused about something?

5. No, I was wrong. It should have been $\displaystyle {\mathbf Z}[2^{1/n}: n=1,2,3,.\ldots]$ with the chain $\displaystyle \left\langle 2^{1/n!} \right\rangle$. Nothing like $\displaystyle {\mathbf Z}[1/2]$ can work since it's a quotient of $\displaystyle {\mathbf Z}[X]$ which is Noetherian.

6. Okay, and that is just a subset of B, using the same idea I had in mind. Thanks.