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Math Help - graded/filtered noetherian algebras

  1. #1
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    graded/filtered noetherian algebras

    Hi, I'm having a bit of bother proving the following statement:


    Let A be a filtered algebra and gr(A) the associated graded algebra. If gr(A) is noetherian without zero-divisors, then so is A.


    Here is my attempt:

    Let A be a filtered algebra with gr(A) noetherian. We have

    gr(A)= \bigoplus_i S_i = \bigoplus_i {F_i(A)}/{F_{i-1}(A)}

    where the F_i(A) are the filtered subspaces of A. Then we have an ascending chain of two-sided ideals (each generated by homogeneous elements) of gr(A)

    I_1 \subset \cdots \subset I_n

    that stabilizes at some n. Under these conditions we may rewrite the above chain of ideals as

    \bigoplus_i I_1 \cap S_i \subset \cdots \subset \bigoplus_i I_n \cap S_i




    I want to show that this new chain is a chain of ideals in the filtered algebra A thus proving A is noetherian. I don't know if this is the right way to go about this. Any suggestions or pointers would be appreciated.

    Sorry if the maths doesn't come out well - this is the first time I've attempted it on this site.
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  2. #2
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    your algebra doesn't have to be both left and right Noetherian. only being left or right Noetherian is good enough.

    so, as you mentioned, let \{F_i\}_{i \geq 0} be a filteration of A and \{I_i\} be some left ideals of A such that I_1 \subseteq I_2 \subseteq \cdots . then gr(I_1) \subseteq gr(I_2) \subseteq \cdots

    is an ascending chain of left ideals of gr(A) and so there exists some positive integer m such that gr(I_m)=gr(I_n) for all n \geq m.

    Claim. for any left ideals of I,J of A if I \subseteq J and gr(I)=gr(J), then I=J.

    if i prove the claim, then your problem is also solved because, since I_m \subseteq I_n for all n \geq m and gr(I_m)=gr(I_n), then by the claim we must have I_m=I_n which proves that A is left Noetherian.

    Proof of the claim. suppose that I \neq J. then there exists (the minimal) positive integer k such that I \cap F_k \subset J \cap F_k. note that here we have

    \subset not \subseteq. the reason for existance of scuh k is that if J \cap F_i = I \cap F_i for all i, then, since A=\bigcup F_i, we'll get the contradiction I=J.

    now let a \in (J \cap F_k) \setminus (I \cap F_k). then a + F_{k-1} \in gr(J)_k=gr(I)_k and thus there exists some b \in I \cap F_k \subset J \cap F_k such that

    a+F_{k-1}=b+F_{k-1}. hence a-b \in J \cap F_{k-1}. but by the minimality of k we have J \cap F_{k-1} = I \cap F_{k-1} and so a-b \in I \cap F_{k-1}.

    therefore a \in I because b \in I and this is the contradiction we needed.
    Last edited by NonCommAlg; January 9th 2011 at 01:07 AM.
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  3. #3
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    of course you also need to prove the second part of your problem, which is quite easy, i.e. if gr(A) is a domain, then A is a domain too.
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