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Math Help - A subring of a Noetherian ring need not be Noetherian-problem

  1. #1
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    A subring of a Noetherian ring need not be Noetherian-problem

    I have a problem with understanding the following solution to the problem:
    A subring of a Noetherian ring is Noetherian? True or false?

    Solution
    let A=k[X,Y] where k is a field, then A is Noetherian by Hilbert's basis theorem.
    Now we want to construct a subring and then show that a sequence of ideals of this subring never becomes stationary thus the subring will not be Noetherian.

    1)We start with letting A_{1}=k+I where I=(X)=AX then A_{1} is a subring of A=k[X,Y]

    as \alpha,\alpha^{\prime}\in k, g,g^{\prime}\in I we have then that
    (\alpha+g)+(\alpha^{\prime}+g^{\prime})\in A_{1} and
    (\alpha+g)\cdot(\alpha^{\prime}+g^{\prime})\in A_{1}

    2) Important observation. If f=\alpha+g\in k+I=A_{1} then f(0,y)=\alpha=(0,0) as g(0,y)=0 and y,y^{2},.....\not\in A_{1}

    3) Hence we may construct an infinite chain of ideals
    (X)\subset (X,XY)\subset (X,XY^2)\subset......
    this chain never stops XY^{k+1}\not\in (X,XY^2,......XY^k)
    as
    XY^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(X,Y)XY^{  i} where all f_{i}\in A_{1}

    \Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,Y)Y^{i}
    \Rightarrow Y^{k+1}=\displaystyle\sum_{i=0}^{k}f_{i}(0,0)Y^{i} (contradition if they were contained).


    Here are my questions and doubts....

    Part 2) How do elements of I=(X)=AX look like?
    are they polynomials generated by element X and then multiplied by any element from the ring? f ex let's take p(X,Y)=3X-4XY\in k[X,Y] then g=X(3X-4XY)=3X^2-4X^2Y\in I Right?

    If we further assume that k=\mathbb{R} we can take for ex \alpha=5 then if f=\alpha+g=5+3X^2-4X^2Y then f(0,y)=5=\alpha=f(0,0) and g(0,y)=0

    but based on what do they conclude that y,y^2,y^3,......\not\in A_{1}??? is it because of the def of A_{1} that it is a ring all polynomials that are all divisible by X??? and these ''alone'' terms with powers of y are not divisible by X

    3) what does the last part of the solution say?

    Could someone explain this solution to me step by step please?
    Thanks
    Last edited by rayman; December 7th 2012 at 01:06 AM.
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  2. #2
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    Re: A subring of a Noetherian ring need not be Noetherian-problem

    your ideal of what AX looks like is essentially correct: every p(x,y) in AX factors as X(g(X,Y)) for some g(X,Y) in k[X,Y].

    so all we do in A1 is add constant terms (from k) to these polynomials. it should be clear for f(X,Y) in A1, f(0,Y) is just an element of k (anything with a Y in it, also has an X with it, so setting X = 0 wipes these all out).

    note that the polynomial Y in k[X,Y] does not have a constant term, so for it to be in A1, it would have to be in I = (X). but Y has no X terms, either (neither does any power of Y).

    now let's look at why XY is not in (X): because if h(X,Y) is in (X), then h(X,Y) = Xf(X,Y) for some f(X,Y) in A1. but this means for h(X,Y) = XY, that f(X,Y) = Y (we can "cancel the X's" since k is a field, so k[X,Y] is an integral domain). and Y isn't in A1.

    the same sort of reasoning shows XY2 isn't in (X,XY): we already know it's not in (X), and if it were in (XY), we could cancel a factor of XY to get Y in A1.

    but maybe XY2 = f1(X,Y)X + f2(X,Y)XY for some f1,f2 in A1? we need to rule this out, too.

    well, first we "cancel the X's" on both sides, and we get:

    Y2 = f1(X,Y) + f2(X,Y)Y

    now the left hand side doesn't have any X's at all that is if: h(X,Y) = Y2, then h(0,Y) = Y2 as well. so we evaluate both sides at (0,Y):

    Y2 = f1(0,Y) + f2(0,Y)Y = a1 + a2Y (for some a1, a2 in k).

    the LHS has degree 2, the RHS has degree 1.
    Thanks from rayman
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  3. #3
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    Re: A subring of a Noetherian ring need not be Noetherian-problem

    thank you very much,that really helped me
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