$X \in \sqrt{{\rm Ann\,( gr\,} M)} \iff \exists m \in \mathbb{N} \text{ such that } \forall j \in \mathbb{Z} ,\; x^mM_j \subset M_{j + mn - 1}$

Question

Let $R$ be a (not necessarily commutative) ring with unity and $M$ a left $R-$module.$\DeclareMathOperator{\Ann}{Ann}\DeclareMathOperator{\gr}{gr}$

Let $(R_i)$ be a $\mathbb{Z}$ filtration on $R$ and $(M_i)$ a filtration on $M$.

Let $\gr R$ be the associated graded ring and $\gr M$ the associated graded module. Recall the action here is given by $(r_i + R_{i-1}, m_j + M_{j-1}) \mapsto r_im_j + M_{i+j-1}$

Take $x \in R_n$, $X := x + R_{n-1}$.

We're asked to show:

$X \in \sqrt{\Ann(\gr M)} \iff \exists m \in \mathbb{N} \text{ such that } \forall j \in \mathbb{Z} \text{ it holds } x^mM_j \subset M_{j + mn - 1}$

The reason I'm posting is because I think we're missing some assumptions:

If we assume that $gr R$ is Noetherian then the direction $ \implies$ is easy because there are primes $P_1,...,P_m$ in $R$ such that $P_1 \cdot ... \cdot P_m \subset \Ann(\gr M)$.

The $\impliedby$ direction is trivial if we assume $\gr R$ is commutative but I can't see why it follows if we only assume $\gr R$ is Noetherian.

Questions:

1. Can this be solved without any extra assumptions, as stated?

2. If you think not, can you solve the $\impliedby$ direction using that $\gr R$ is Noetherian alone?

Answer 1

Let $I=\mathrm{Ann}(\mathrm{gr}M)$, so that $x\in I\cap R_n$ if and only if $xM_j\subset M_{j+n-1}$. Also, $\sqrt I$ is the intersection of all prime ideals containing $I$, so $x\in\sqrt I$ if and only if $x+I$ is strongly nilpotent in $R/I$. That is, every sequence $x=x_0,x_1,x_2,\ldots$ with $x_{i+1}\in x_iRx_i+I$ eventually lies in $I$. Clearly every strongly nilpotent element is nilpotent, and so the implication $\Rightarrow$ is always true. The converse is not true, since nilpotent elements are in general not strongly nilpotent.