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Let $R$ be Noetherian, $I\subseteq R$ an ideal, $S=1+I$, and $M$ a finitely generated $R$-module. ThenIf $(1+a)m=0$ with $a\in I$, then $m=(-a{)}^{j}m\in I^{j}M$ for every $j$. Conversely, apply the Artin-Rees lemma to $Rm\subseteq M$. For sufficiently large $c$ it givesso $m=am$ for some $a\in I$ and $(1-a)m=0$.

A local ring that is also an integral domain. Multiplication by any nonzero element is injective. This permits cancellation in Artin-Rees lemma length-growth arguments.

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 height of a prime ideal is the supremum of the lengths of strict chains of prime ideals ending at $P$:The last equality follows from the prime ideal correspondence for localization. Here is a length-growth proof of the Krull height theorem that makes the bound explicit.

We first establish a prime-chain lower bound for local length. For a Noetherian local ring $(B,\mathfrak{m})$ admitting a prime chain of length $d$, put $H_B(t)=\mathrm{length}(B/{\mathfrak{m}}^t)$. Thenfor some $c>0$. For $d=0$, use $H_B(t)\ge 1$. For $d>0$, quotient by the first prime ideal of the chain; this only decreases $H_B$ and reduces us to a local integral domain $B$ withTake $0\ne a\in P_1$ and put $C=B/(a)$. It has a prime chain of length $d-1$, so induction bounds $H_C(t)$ below by a positive multiple of $t^{d-1}$.

The Artin-Rees lemma applied to $(a)\subseteq B$ gives, for a fixed integer $c_0\ge 1$,Because $a$ is a non-zero-divisor, the first term in the exact sequencehas length at least $H_B(t-c_0)$. HenceIterate this inequality about $t/(2c_0)$ times, while its arguments remain at least $t/2$. This gives the desired positive multiple of $t^d$. The particular Artin-Rees lemma inclusion used here follows directly from finite generation: the Rees algebra ${⨁}_{j\ge 0}{\mathfrak{m}}^j{t}^j$ is Noetherian, by the Hilbert basis theorem, and its graded submodule ${⨁}_j((a)\cap {\mathfrak{m}}^j)t^j$ has finitely many homogeneous generators. A bound on their degrees gives $c_0$.

Now localize at the given minimal prime ideal $P$. Write $B=R_P$, $\mathfrak{m}=P{R}_P$, and $J=I{R}_P$. The only prime ideal of $B$ containing $J$ is $\mathfrak{m}$, so $\sqrt{J}=\mathfrak{m}$. Since $\mathfrak{m}$ is finitely generated, some ${\mathfrak{m}}^q\subseteq J$. Thus $B/J$ has finite module length, say $L$. The ideal $J$ is generated by at most $n$ elements.

For $n>0$, products of these generators give a surjection from $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $B/J$ onto $J^j/J^{j+1}$. Summing the lengths gives the generator bound for primary-ideal lengthSince $J\subseteq \mathfrak{m}$, we have $J^t\subseteq {\mathfrak{m}}^t$, and thereforeThe prime-chain lower bound for local length now excludes any chain of length greater than $n$. ConsequentlyFor $n=0$, minimality over the zero ideal says that $P$ is a minimal prime ideal, so its height is zero. This also handles that boundary case.

A local ring is a nonzero ring with one maximal ideal $\mathfrak{m}$. Write $\kappa =R/\mathfrak{m}$ for its residue field. Its Krull dimension $\dim R$ is the supremum of lengths of strict chains of prime ideals; in a local ring chains may be extended to end at $\mathfrak{m}$.

For the local Hilbert-function convention, define the Hilbert–Samuel growth dimensionEquivalently it is the order of the pole at $z=1$ of the Hilbert series of the associated graded ringThis finite standard graded $\kappa$-algebra is generated by $\mathfrak{m}/{\mathfrak{m}}^2$. The Hilbert-Serre theorem makes the cumulative Hilbert function eventually polynomial, which establishes the definition. For an Artinian local ring, the polynomial is constant and nonzero, so $d(R)=0$.

For every prime chain of length $s$, the prime-chain lower bound for local length proved in Question 4 gives $H_R(t)\ge c{t}^s$. A polynomial of degree $d(R)$ cannot satisfy this when $s>d(R)$. Thus the requested inequality isFor clarity, the embedding dimension is the different invariantThe equality follows from the Nakayama lemma. Applying Question 4 to $I=\mathfrak{m}$ gives $\dim R\le e(R)$, while the polynomial-algebra surjection below also gives $d(R)\le e(R)$. In particular, growth dimension and embedding dimension should be kept distinct.

A regular local ring is a Noetherian local ring with $e(R)=\dim R$. Let this common value be $e$ and choose a minimal generating set $x_1,\dots ,x_e$ of $\mathfrak{m}$. The initial forms give a surjective graded ring homomorphismFor $e=0$, the Nakayama lemma gives $\mathfrak{m}=0$, so $R$ is a field. Assume $e>0$. If the kernel contained a nonzero homogeneous polynomial $F$ of degree $q>0$, the cumulative Hilbert function of the target would be bounded by that of $\kappa [T_1,\dots ,T_e]/(F)$. Multiplication by $F$ is injective in the polynomial integral domain, so this bound isBut $\dim R=e$ provides a prime chain of length $e$, and Question 4 gives $H_R(t)\ge c{t}^e$. This is a contradiction. Hence the associated graded ring of a regular local ring isan integral domain.

Finally the Krull intersection theorem gives ${\bigcap }_j{\mathfrak{m}}^j=0$. One can see the needed separatedness directly: for the finitely generated ideal $K={\bigcap }_j{\mathfrak{m}}^j$, the Artin-Rees lemma gives $K=\mathfrak{m}K$, and the Nakayama lemma gives $K=0$. Therefore each nonzero $a\in R$ has finite $\mathfrak{m}$-adic order $r$, with nonzero initial form in ${\mathfrak{m}}^r/{\mathfrak{m}}^{r+1}$. For nonzero $a,b$, their initial forms have nonzero product in the graded integral domain. It follows that $ab\ne 0$, and indeed its order is the sum of their orders. Thus $R$ is an integral domain.

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 $m$.

The set $S=1+I$ is a multiplicative subset, since $(1+a)(1+b)=1+(a+b+ab)$. The localization of a ring $S^{-1}R$ consists of fractions $r/s$, whereSimilarly, the localization of a module $S^{-1}M$ consists of fractions $m/s$, with the analogous equivalence relation. The canonical maps areThe Localization of a Noetherian ring is Noetherian. Since a finitely generated module over a Noetherian ring is a Noetherian module, localizing a finite generating set shows that $S^{-1}M$ is Noetherian over $S^{-1}R$.

The prime ideal correspondence for localization identifies $\mathrm{Spec}{S}^{-1}R$ with the primes $P\in \mathrm{Spec}R$ disjoint from $S$. Hereso the spectrum of localization away from one plus an ideal is

An element $m$ maps to zero exactly when $(1+a)m=0$ for some $a\in I$. Such an equation gives $m=(-a{)}^{j}m\in I^{j}M$ for every $j$, proving one inclusion. Conversely, suppose $m\in {\bigcap }_{j\ge 1}{I}^{j}M$. The Artin-Rees lemma applied to $Rm\subseteq M$ says that for some $c$,Taking $n=c+1$ gives $m\in I(Rm)$, so $m=am$ for some $a\in I$. Then $(1-a)m=0$ and $1-a\in S$. Thus

For failure without Noetherianity, takeThe strict chain $(t)⊊(t^{1/2})⊊(t^{1/4})⊊\cdots$ shows that $R$ is not Noetherian. Since $t^q=(t^{q/2}{)}^2$, one has $I^2=I$, and hence ${\bigcap }_j{I}^j=I\ne 0$. But $R$ is an integral domain, so its localization map into $(1+I{)}^{-1}R$ is injective. This is the non-Noetherian failure of the intersection formula for localization.

If the filtration is stable from degree $\ell$, then $M^{\ast }$ is generated over $R^{\ast }$ by finite generating sets for $M_0,\dots ,M_{\ell }$. Conversely, let homogeneous elements of degrees at most $\ell$ generate $M^{\ast }$. In every degree $n>\ell$, each expression for an element of $M_n$ uses a positive-degree coefficient from $R^{\ast }$, so $M_n=I{M}_{n-1}$. Hence finite generation is equivalent to stability.

For $M^{\prime }\subseteq M$, takeThen $M^{\ast }$ is a graded submodule of the finite Rees module $⨁I^{n}M$. Since $R$ is Noetherian and $I$ is finitely generated, $R^{\ast }$ is Noetherian; hence $M^{\ast }$ is finite and the filtration is stable. Therefore, for some $\ell$ and all $n\ge \ell$,This is the Artin-Rees lemma.

Work modulo $I$. Putand let $A$ be the image of $I^{\prime }/I$ in $C$. The separating condition $I^{\prime }=I^{\prime }\cap (I+J)$ modulo $I$ says that $I^{\prime }/I\to C$ is injective, so we may regard $A$ as a submodule of $C$.

Since $r\in \sqrt{(I:I^{\prime })}$, some power $r^q$ annihilates $A$. Apply the Artin-Rees lemma to $A\subseteq C$ and the principal ideal $(r)$. There is $s$ such that for every $m\ge s$,For $m\ge s+q$, the right side is zero. Pulling the equality $A\cap r^{m}C=0$ back to $R$ givesas required.

A prime chain of length $s$ in a Noetherian local ring forces its maximal-ideal length function to be at least $c{t}^s$ for large $t$. Quotient by the initial prime to reduce to a domain, choose a nonzero $a$ in the next prime, and use the Artin-Rees lemma to obtain $H_R(t)\ge H_R(t-c_0)+H_{R/(a)}(t)$. Induction on chain length and summing about $t/(2c_0)$ terms prove the bound. This yields $\dim R\le d(R)$ without assuming the Krull height theorem.

The Rees module of an $R$-module $M$ with respect to an ideal $I$ is ${\mathcal{R}}_I(M)={⨁}_{j\ge 0}{I}^{j}M{t}^j$, a graded module over the Rees ring $\mathcal{R}(I)$. Generators of $M$ in degree zero generate this module. For $N\subseteq M$, the graded submodule ${⨁}_j(I^{j}M\cap N)t^j$ is finitely generated when $R$ is a Noetherian ring; a bound on its generators' degrees yields the Artin-Rees lemma.