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.

2. 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.

3. 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.