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Let $I=(a_1,\dots ,a_s)$ and form the Rees algebra $\mathcal{R}={⨁}_{j\ge 0}{I}^j{t}^j$. It is generated over $R$ by $a_{1}t,\dots ,a_{s}t$, so the Hilbert basis theorem makes it Noetherian. The Rees module $\mathcal{M}={⨁}_{j\ge 0}{I}^{j}M{t}^j$ is finite over it, generated by finitely many generators of the module $M$ in degree zero. Hence its graded submodulehas finitely many homogeneous generators, all in degrees at most some $k$.

For $i\ge k$, each degree-$d$ generator contributes a multiple from $I^{i-d}(N\cap I^{d}M)$. This is contained in $I^{i-k}(N\cap I^{k}M)$, because multiplication by $I^{k-d}$ puts its coefficient both in $N$ and in $I^{k}M$. Thus $N\cap I^{i}M\subseteq I^{i-k}(N\cap I^{k}M)$. The reverse inclusion follows directly by multiplication: it stays in $N$ and in $I^{i}M$. ConsequentlyThis is the Artin-Rees lemma, proved through finite graded generation rather than an assumption that the intersection filtration initially has that form.

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 $c\ge 0$ such thatThus 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:Since $R$ is a Noetherian ring, write $I=(a_1,\dots ,a_s)$. Then $\mathcal{R}(I)=R[a_{1}t,\dots ,a_{s}t]$, 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 submoduleis 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 givesThe same generators show that the final sum is exactly $I^{c}M\cap N$. This proves the Artin-Rees lemma. If $L=0$, one may simply take $c=0$.

Now letIt 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 $I^{j}M$ (including $I^{0}M=M$), its conclusion reduces toWrite generators as $y_1,\dots ,y_h$. 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 thatThe 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 $m=(-r{)}^{j}m\in I^{j}M$ for every $j\ge 1$. Hence the exact description isIn 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 $D=\mathbb{Z}[1/2{]}_{\ge 0}$ and use the semigroup algebraIts elements are finite sums of formal monomials $t^q$, with multiplication $t^q{t}^{q^{\prime }}=t^{q+q^{\prime }}$. 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, sinceIndeed 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, equivalentlyIt 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 $j\ge 1$. TakeFor every $r\in I$, the constant term of $1+r$ is $1$, so $1+r\ne 0$. Since $R$ is an integral domain and $t\ne 0$, we have $(1+r)t\ne 0$. This gives all three requested objects and proves that no element of the required form annihilates $m$.