= Solution
The <Artin-Rees lemma> states that, for a <Noetherian ring> $R$, an <ideal> $I$, a <finitely generated module> $M$, and a <submodule> $N\subseteq M$, there is $c\ge0$ such that
$$
\boxed{I^jM\cap N=I^{j-c}(I^cM\cap N)\quad\text{for every }j\ge c.}
$$
Thus the filtration on $N$ induced from the <I-adic filtration> of $M$ is eventually determined by one of its terms.
For the proof, form the <Rees ring> and its <Rees module>:
$$
\mathcal R(I)=\bigoplus_{j\ge0}I^jt^j\subseteq R[t],\qquad
\mathcal R_I(M)=\bigoplus_{j\ge0}I^jMt^j.
$$
Since $R$ is a <Noetherian ring>, write $I=(a_1,\ldots,a_s)$. Then $\mathcal R(I)=R[a_1t,\ldots,a_st]$, so it is a <Noetherian ring> by the <Hilbert basis theorem>. Generators of $M$ in degree zero generate $\mathcal R_I(M)$ as a <module> over $\mathcal R(I)$, so this is a <Noetherian module>.
The graded <submodule>
$$
L=\bigoplus_{j\ge0}(I^jM\cap N)t^j
$$
is therefore finitely generated. Choose homogeneous generators $z_\alpha t^{d_\alpha}$ with $d_\alpha\le c$ for some $c$. For $j\ge c$, taking degree-$j$ components gives
$$
I^jM\cap N=\sum_\alpha I^{j-d_\alpha}z_\alpha
=I^{j-c}\sum_\alpha I^{c-d_\alpha}z_\alpha.
$$
The same generators show that the final sum is exactly $I^cM\cap N$. This proves the <Artin-Rees lemma>. If $L=0$, one may simply take $c=0$.
Now let
$$
K=\bigcap_{j\ge1}I^jM.
$$
It is a <submodule> of the <Noetherian module> $M$, hence a <finitely generated module>. Apply the <Artin-Rees lemma> with $N=K$ and $j=c+1$. Since $K$ is contained in every $I^jM$ (including $I^0M=M$), its conclusion reduces to
$$
K=IK.
$$
Write generators as $y_1,\ldots,y_h$. There is a matrix $A=(a_{ij})$ with entries in $I$ such that $y_i=\sum_j a_{ij}y_j$. Multiplication by the adjugate of $I_h-A$ shows that
$$
\det(I_h-A)y_i=0\quad\text{for all }i.
$$
The <determinant> is congruent to $1$ modulo $I$, so it is $1+r$ for some $r\in I$. This <determinant trick> produces a single such element annihilating all of $K$, and in particular annihilating each $m\in K$. If $K=0$, use $r=0$.
Conversely, if $(1+r)m=0$ with $r\in I$, then $m=-rm$. Iterating gives $m=(-r)^jm\in I^jM$ for every $j\ge1$. Hence the exact description is
$$
\boxed{\bigcap_{j\ge1}I^jM=\{m\in M:\text{some }r\in I\text{ satisfies }(1+r)m=0\}.}
$$
In a <local ring> with $I$ contained in the <maximal ideal>, every $1+r$ is a <unit>, so this also yields the usual vanishing form of the <Krull intersection theorem>.
For a failure without the <Noetherian ring> hypothesis, let $D=\mathbb Z[1/2]_{\ge0}$ and use the <semigroup algebra>
$$
R=k[t^q:q\in D]=\bigcup_{h\ge0}k[t^{1/2^h}].
$$
Its elements are finite sums of formal <monomials> $t^q$, with multiplication $t^qt^{q'}=t^{q+q'}$. Each ring in the union is a <polynomial ring> and the inclusions are injective, so $R$ is an <integral domain>. It is not a <Noetherian ring>, since
$$
(t)\subsetneq(t^{1/2})\subsetneq(t^{1/4})\subsetneq\cdots.
$$
Indeed the generator on the left is the square of the next generator, while division in the opposite direction would require a negative exponent, which is unavailable in $R$.
Let $I$ be the <ideal> of elements with zero constant term, equivalently
$$
I=(t^q:q\in D,\ q>0).
$$
It is proper because taking the constant term gives $R/I\cong k$. Every generator $t^q$ is $(t^{q/2})^2$, with $t^{q/2}\in I$, so $I^2=I$. Thus this is a nonzero <idempotent ideal>, with $I^j=I$ for every $j\ge1$. Take
$$
\boxed{m=t\in\bigcap_{j\ge1}I^j.}
$$
For every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne0$. Since $R$ is an <integral domain> and $t\ne0$, we have $(1+r)t\ne0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.
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