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Thread: 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

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

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

    $\displaystyle I_1 \subset \cdots \subset I_n $

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

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




    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 $\displaystyle \{F_i\}_{i \geq 0}$ be a filteration of $\displaystyle A$ and $\displaystyle \{I_i\}$ be some left ideals of $\displaystyle A$ such that $\displaystyle I_1 \subseteq I_2 \subseteq \cdots $. then $\displaystyle gr(I_1) \subseteq gr(I_2) \subseteq \cdots $

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

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

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

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

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

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

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

    therefore $\displaystyle a \in I$ because $\displaystyle b \in I$ and this is the contradiction we needed.
    Last edited by NonCommAlg; Jan 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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