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Math Help - Non noetherian rings

  1. #1
    DMT
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    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?
    Last edited by DMT; May 12th 2006 at 12:33 PM.
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  2. #2
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    Try {\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 \left\langle 1/2 \right\rangle \subset \left\langle (1/2)^2 \right\rangle \subset \left\langle (1/2)^3 \right\rangle  \subset \cdots .
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  3. #3
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    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
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  4. #4
    DMT
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    Quote Originally Posted by rgep
    Try {\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 \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?
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  5. #5
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    No, I was wrong. It should have been {\mathbf Z}[2^{1/n}: n=1,2,3,.\ldots] with the chain \left\langle 2^{1/n!} \right\rangle. Nothing like {\mathbf Z}[1/2] can work since it's a quotient of {\mathbf Z}[X] which is Noetherian.
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  6. #6
    DMT
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    Okay, and that is just a subset of B, using the same idea I had in mind. Thanks.
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