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

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

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- May 12th 2006, 12:30 PMDMTNon 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? - May 12th 2006, 01:40 PMrgep
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 $.

- May 12th 2006, 01:42 PMBubbleBrain_103A 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 - May 12th 2006, 02:20 PMDMTQuote:

Originally Posted by**rgep**

- May 12th 2006, 10:46 PMrgep
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.

- May 13th 2006, 01:18 AMDMT
Okay, and that is just a subset of B, using the same idea I had in mind. Thanks.