graded/filtered noetherian algebras

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January 8th 2011, 09:31 PM
hp32

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.
January 9th 2011, 12:04 AM
NonCommAlg

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.
January 9th 2011, 12:16 AM
NonCommAlg

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.