Rees Modules Artin Rees and Hilbert Samuel Theory

Induct on $r$. If $r=0$, the standard grading makes $S=A$ concentrated in degree zero. Choose a finite homogeneous generating list for $M$ by splitting any finite generating list into its finitely many homogeneous components. Only the finitely many degrees of those generators occur in $M$, and each such degree piece is finite over $A$, hence has finite length by [L3]. Therefore ${\mathrm{HS}}_M(t)$ is a Laurent polynomial and its coefficients are zero for all sufficiently large $n$. Both conclusions hold in this base case.

Assume $r>0$ and the assertion for standard graded algebras presented using $r-1$ variables. By [L3] and [L4], $S$ is Noetherian; hence the kernel of multiplication by $x_r$ on the finite module $M(-1)$ is finite, as is its cokernel. Write $S^{\mathrm{\prime }}=A[x_1,\dots ,x_{r-1}]\mathrm{/}(J\cap A[x_1,\dots ,x_{r-1}])$. Multiplication by the degree-one class of $x_r$ gives an exact sequence of graded $S$-modules $0\to K\to M(-1)\stackrel{x_r}{\to }M\to C\to 0.$ Both $K$ and $C$ are annihilated by $x_r$ and are therefore finite graded modules over the $r-1$-variable algebra $S^{\mathrm{\prime }}$: split their finite $S$-generating lists into homogeneous components, and the same lists generate after the action factors through $S\mathrm{/}(x_r)$, a quotient of $S^{\mathrm{\prime }}$.

Every degree piece of $M$, $K$, and $C$ is a finite $A$-module: finite homogeneous generators and finitely many monomials of any fixed degree suffice. By [L3], it has finite length. Taking degree-$n$ pieces in the exact sequence of step 1.2 and using [L2] yields ${\mathrm{\ell }}_A(C_n)-{\mathrm{\ell }}_A(K_n)={\mathrm{\ell }}_A(M_n)-{\mathrm{\ell }}_A(M_{n-1})$ for every $n$. In Hilbert-series form this is $(1-t){\mathrm{HS}}_M(t)={\mathrm{HS}}_C(t)-{\mathrm{HS}}_K(t)$ by [L1].

The induction hypothesis applies to the finite graded $S^{\mathrm{\prime }}$-modules $K$ and $C$ from step 1.2. Their series have Laurent-polynomial numerators over $(1-t{)}^{r-1}$. Equation 2.1 therefore writes ${\mathrm{HS}}_M(t)$ with a Laurent-polynomial numerator over $(1-t{)}^r$.

Write the numerator from step 3.1 as $p(t)={\sum }_{j=a}^b{c}_j{t}^j$ with finite integers $c_j$. For $r>0$, the coefficient of $t^n$ in $t^j(1-t{)}^{-r}$ is $\left(\genfrac{}{}{0px}{}{n-j+r-1}{r-1}\right)$ for $n\ge j$. For all $n\ge b$, the Hilbert function is therefore ${\sum }_{j=a}^b{c}_j\left(\genfrac{}{}{0px}{}{n-j+r-1}{r-1}\right)$, a polynomial in $n$ with rational coefficients.

The base case is step 1.1, and steps 1.2 through 4.1 prove the induction step. Thus both claims hold for every finite number of variables.

If the filtration is $I$-stable from some index $c$ onward, then $\mathcal{R}(M_{\bullet })={\sum }_{n=0}^c\mathcal{R}(I)\cdot (M_n{t}^n),$ because for $n>c$ every element of $M_n{t}^n$ is a product of an element of $I^{n-c}{t}^{n-c}$ with one of $M_c{t}^c$. Since $R$ is Noetherian and $M$ is finite, [L2] makes each submodule $M_n\subseteq M$ finitely generated. Therefore finitely many homogeneous elements in degrees at most $c$ generate $\mathcal{R}(M_{\bullet })$ over $\mathcal{R}(I)$.

Conversely, suppose $\mathcal{R}(M_{\bullet })$ is generated over $\mathcal{R}(I)$ by homogeneous elements lying in degrees at most $c$. For $n\ge c$, every element of $M_{n+1}{t}^{n+1}$ is therefore a sum of products $(I^{n+1-d}{t}^{n+1-d})(m_d{t}^d)$ with $d\le c$, hence lies in $I{M}_n{t}^{n+1}$. Thus $M_{n+1}\subseteq I{M}_n$. The reverse inclusion is part of the filtration hypothesis, so $M_{n+1}=I{M}_n$ for all $n\ge c$.

By [L3], choose generators $I=(f_1,\dots ,f_s)$. Sending $X_i$ to $f_{i}t$ defines a surjective graded map $R[X_1,\dots ,X_s]\twoheadrightarrow \mathcal{R}(I)\mathrm{.}$ By [L4] its source is Noetherian, so its quotient $\mathcal{R}(I)$ is Noetherian.

Steps 1.1 and 1.2 prove the equivalence, and step 1.3 proves that $\mathcal{R}(I)$ is Noetherian.

The $I$-adic filtration on $M$ is already stable, since $I(I^{n}M)=I^{n+1}M$ for every $n\ge 0$. Hence [L1] makes ${\mathcal{R}}_I(M):={⨁}_{n\ge 0}{I}^{n}M{t}^n$ a finite module over the Noetherian ring $\mathcal{R}(I)$.

The induced filtration $N_n:=I^{n}M\cap N$ defines a graded submodule ${\mathcal{R}}_N:={⨁}_{n\ge 0}{N}_n{t}^n\subseteq {\mathcal{R}}_I(M)\mathrm{.}$ The module ${\mathcal{R}}_I(M)$ is finite over the Noetherian ring $\mathcal{R}(I)$ by step 1.1 and [L1], hence is Noetherian by [L2]. Therefore its submodule ${\mathcal{R}}_N$ is finite.

Applying the stability direction of [L1] to the finite Rees module established in step 2.1 yields an index $c$ with $N_n=I^{n-c}{N}_c(n\ge c)\mathrm{.}$ Since $N_n=I^{n}M\cap N$ and $N_c=I^{c}M\cap N$, this is exactly the displayed Artin-Rees equality.

Therefore the required constant $c$ exists.

For every $n$, one always has $I^{n}N\subseteq I^{n}M\cap N$, since $N\subseteq M$.

Choose $c$ as in [L1]. Then for $n\ge c$, $I^{n}M\cap N=I^{n-c}(I^{c}M\cap N)\subseteq I^{n-c}N$ because $I^{c}M\cap N\subseteq N$.

Combining the preceding steps gives the two-sided eventual inclusion, hence the two filtrations are equivalent.

Since $K\subseteq I^{n}M$ for every $n$, Artin-Rees gives some $c$ such that for all $n\ge c$, $K=I^{n}M\cap K=I^{n-c}(I^{c}M\cap K)=I^{n-c}K\mathrm{.}$ In particular $IK=K$.

Conversely, if $(1-a)m=0$ for some $a\in I$, then $m=am\in IM$. Iterating gives $m=a^{n}m\in I^{n}M$ for every $n\ge 0$, hence $m\in K$.

Steps 2.1 and 1.2 identify $K$ with the set of $(1-a)$-torsion elements claimed in part 1.

The submodule $K\subseteq M$ is finite by [L4]. Applying [L2] to $K$ and the equality $IK=K$ from step 1.1 gives $a\in I$ with $(1-a)K=0.$ So every element of $K$ satisfies the displayed torsion condition.

Assume now $I\subseteq J(R)$. For any $m\in K$, step 2.1 gives some $a\in I\subseteq J(R)$ with $(1-a)m=0$. By [L3], $1-a$ is a unit, so multiplying by its inverse gives $m=0$. Thus $K=0$.

Therefore both stated conclusions hold.

Every simple module over the local ring $R$ is $R\mathrm{/}\mathfrak{m}$: a nonzero element generates a simple module, whose annihilator is a maximal ideal, necessarily $\mathfrak{m}$. Along a composition series of the finite-length module $M\mathrm{/}IM$, each multiplication by $\mathfrak{m}$ moves into the preceding term. Thus ${\mathfrak{m}}^c(M\mathrm{/}IM)=0$ for some $c\ge 1$, or ${\mathfrak{m}}^{c}M\subseteq IM$. Multiplying by $I^n$ gives ${\mathfrak{m}}^c{I}^{n}M\subseteq I^{n+1}M$ for every $n\ge 0$. The quotient $A:=R\mathrm{/}{\mathfrak{m}}^c$ is Artinian: $\mathfrak{m}$ is finitely generated by [L3], so each factor ${\mathfrak{m}}^i\mathrm{/}{\mathfrak{m}}^{i+1}$ for $0\le i<c$ is a finite-dimensional vector space over $R\mathrm{/}\mathfrak{m}$; their finite filtration gives $A$ finite length as an $R$-module. Each graded piece $I^{n}M\mathrm{/}{I}^{n+1}M$ is naturally an $A$-module.

By [L3], choose generators $x_1,\dots ,x_r$ of $I$, and choose generators $m_1,\dots ,m_s$ of $M$. The classes of the $m_j$ in degree $0$ generate ${\mathrm{gr}}_I(M)$ over the standard graded $A$-algebra $A[T_1,\dots ,T_r]$, where $T_i$ acts by multiplication with the class of $x_i$. Indeed every class in $I^{n}M\mathrm{/}{I}^{n+1}M$ is represented by a finite sum of monomials $x_{i_1}\cdots x_{i_n}{m}_j$.

Length over $R$ and over $A=R\mathrm{/}{\mathfrak{m}}^c$ agree on each module $I^{n}M\mathrm{/}{I}^{n+1}M$, because that quotient is annihilated by ${\mathfrak{m}}^c$ and has the same submodules in either category. Applying [L2] to the finite graded $A[T_1,\dots ,T_r]$-module ${\mathrm{gr}}_I(M)$, the function $n⟼{\mathrm{\ell }}_R(I^{n}M\mathrm{/}{I}^{n+1}M)$ agrees for large $n$ with a polynomial $Q(n)$. Equivalently, the Hilbert series of ${\mathrm{gr}}_I(M)$ is rational with denominator a power of $(1-t)$.

By [L1], the Hilbert-Samuel function is the cumulative sum of these graded-piece lengths: ${\chi }_{I,M}(n)={\sum }_{j=0}^n{\phi }_{I,M}(j)\mathrm{.}$ The rational Hilbert series in [L2] has an integral Laurent numerator and denominator $(1-t{)}^r$; cumulative summation divides it by $(1-t)$. Its large-degree coefficients are therefore integer linear combinations of shifted binomial polynomials $\left(\genfrac{}{}{0px}{}{n-k+r}{r}\right)$. The finite-difference identity for integer-valued polynomials expresses each such shift as an integer linear combination of $\left(\genfrac{}{}{0px}{}{n+j}{j}\right)$, $0\le j\le r$. Thus ${\chi }_{I,M}(n)$ is eventually a polynomial in the claimed binomial form.

Hence the Hilbert-Samuel polynomial exists.

Put $A=R\mathrm{/}{\mathrm{Ann}}_R(M)$ and let $J=IA$. Then $A$ is Noetherian local, $M$ is a faithful finite $A$-module, and $\mathrm{Spec}(A)={\mathrm{Supp}}_R(M)$. If $m_1,\dots ,m_s$ generate $M$, there is a surjection $A^s\twoheadrightarrow M$ and a faithful injection $A\hookrightarrow M^s$, $a\mapsto (a{m}_1,\dots ,a{m}_s)$. The surjection gives ${\chi }_{J,M}(n)\le s{\chi }_{J,A}(n)$. Applying [L3] to the injection gives a bounded shift $c$ and the reverse estimate ${\chi }_{J,A}(n-c)\le s{\chi }_{J,M}(n)$ for large $n$. Hence $P_{J,A}$ and $P_{I,M}$ have the same degree. It remains to prove the theorem for the ring $A$.

Let $r=\mathrm{deg}{P}_{J,A}$ and let $d$ be the least number of generators of an ideal of definition of $A$; [L1] identifies $d$ with $\dim A$. Choose an ideal of definition $Q=(x_1,\dots ,x_d)$. The ideals $J$ and $Q$ are finite by [L6] and have the same radical ${\mathfrak{m}}_A$. Raising finite generating sets to suitable powers and expanding products therefore gives positive integers $u,v$ with $J^u\subseteq Q$ and $Q^v\subseteq J$. The resulting linear reindexing inequalities between ${\chi }_{J,A}$ and ${\chi }_{Q,A}$ show that their eventual polynomials have the same degree. If $d=0$, then $Q=0$ is an ideal of definition, so $A$ has finite length and $r=\mathrm{deg}{P}_{Q,A}=0$. If $d>0$, every $Q^n\mathrm{/}{Q}^{n+1}$ is a quotient of a direct sum of $\left(\genfrac{}{}{0px}{}{n+d-1}{d-1}\right)$ copies of $A\mathrm{/}Q$, indexed by the degree-$n$ monomials in the $x_i$. Its length is therefore bounded by a polynomial of degree $d-1$, and summing the graded-piece lengths gives $r=\mathrm{deg}{P}_{Q,A}\le d$.

We prove $\dim A\le r$ by induction on $r$. If $r=0$, the increasing integer sequence ${\mathrm{\ell }}_A(A\mathrm{/}{J}^{n+1})$ is eventually constant, so $J^n\mathrm{/}{J}^{n+1}=0$ for all large $n$. Thus the finite ideal $J^n$ from [L6] satisfies $J(J^n)=J^n$. By [L5], some $a\in J$ has $(1-a)J^n=0$; since $a\in J\subseteq {\mathfrak{m}}_A$, [L7] makes $1-a$ a unit, and hence $J^n=0$. Because $J$ is an ideal of definition, $\sqrt{J}={\mathfrak{m}}_A$; nilpotence of $J$ then makes every prime equal to ${\mathfrak{m}}_A$, so $\dim A=0$.

Assume $r>0$ and take a strict chain ${\mathfrak{p}}_0\subset {\mathfrak{p}}_1\subset \cdots \subset {\mathfrak{p}}_e={\mathfrak{m}}_A$. If $e=0$, there is nothing to prove, so assume $e>0$. Put $B=A\mathrm{/}{\mathfrak{p}}_0$. The quotient maps $A\mathrm{/}{J}^{n+1}A\twoheadrightarrow B\mathrm{/}{J}^{n+1}B$ give ${\chi }_{J,B}(n)\le {\chi }_{J,A}(n)$, hence $s:=\mathrm{deg}{P}_{J,B}\le r$. Choose $x\in {\mathfrak{p}}_1\setminus {\mathfrak{p}}_0$ and write $\bar{x}$ for its nonzero image in the domain $B$. Multiplication by $\bar{x}$ is injective, so $0\to B\stackrel{\bar{x}}{\to }B\to C:=B\mathrm{/}\bar{x}B\to 0$ is short exact; the image of $J$ is an ideal of definition in both $B$ and $C$. Applying [L2] at the maximum of $s$ and $\mathrm{deg}{P}_{J,C}$ cancels the two degree-$s$ contributions from $B$; if $\mathrm{deg}{P}_{J,C}\ge s$, it would force the nonzero leading coefficient of $P_{J,C}$ to vanish. Hence $\mathrm{deg}{P}_{J,C}<s\le r$. The images of ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_e$ give a strict prime chain of length $e-1$ in $C$. The induction hypothesis applied to $C$ gives $e-1\le \dim C\le \mathrm{deg}{P}_{J,C}\le r-1,$ and therefore $e\le r$. Since the chain was arbitrary, $\dim A\le r$.

Steps 1.2 and 1.4 give $r=d=\dim A$, and step 1.1 transfers this equality to $M$. Therefore $\mathrm{deg}{P}_{I,M}=\dim \mathrm{Supp}(M)$.

Let $A:=R\mathrm{/}{\mathrm{Ann}}_R(M)$ and $\mathfrak{n}:=\mathfrak{m}\mathrm{/}{\mathrm{Ann}}_R(M)$. The ring $A$ is nonzero, Noetherian and local; $M$ is a faithful finite $A$-module. By [L5], ${\mathrm{Supp}}_A(M)=\mathrm{Spec}(A)$, and prime correspondence identifies this space with ${\mathrm{Supp}}_R(M)$, so $d=\dim A={\mathrm{ht}}_A(\mathfrak{n})$. The ideal $\mathfrak{n}$ has a finite generating list by [L3]; since it is maximal, it is minimal over its own generating ideal. Under the assumed AC, [L4] bounds its height by that list length. Thus $d<\mathrm{\infty }$ and [L1] applies to $A$.

A nonzero finite $A$-module $Q$ has finite length exactly when its support is $\{\mathfrak{n}\}$. In one direction, a finite composition series has simple factors $A\mathrm{/}\mathfrak{n}$: for a nonzero element of a simple factor, its annihilator is maximal, hence is the unique maximal ideal $\mathfrak{n}$. Thus ${\mathfrak{n}}^{s}Q=0$ for the series length $s$, and [L5] gives support contained in $\{\mathfrak{n}\}$; nonzero $Q$ has $\mathfrak{n}$ in its support by [L6]. Conversely, if support is $\{\mathfrak{n}\}$, [L5] gives $\sqrt{{\mathrm{Ann}}_A(Q)}=\mathfrak{n}$. Choose finite generators $t_1,\dots ,t_u$ of $\mathfrak{n}$ by [L3]. Each $t_i$ has a positive power annihilating $Q$, so some power ${\mathfrak{n}}^N$ annihilates $Q$. The finite filtration $Q\supseteq \mathfrak{n}Q\supseteq \cdots \supseteq {\mathfrak{n}}^{N}Q=0$ has finite $A\mathrm{/}\mathfrak{n}$-vector-space quotients: each ${\mathfrak{n}}^{i}Q$ is generated by products of finite generating lists for $\mathfrak{n}$ and $Q$. Refining each vector-space quotient by a basis flag gives a finite composition series.

Fix an ideal $J\subseteq \mathfrak{m}$ and put $Q=M\mathrm{/}JM$. By [L6], $Q\mathrm{\ne }0$. For each prime $\mathfrak{q}$ of $A$, if $JA⊈\mathfrak{q}$ then $Q_{\mathfrak{q}}=0$. If $JA\subseteq \mathfrak{q}$, then $M_{\mathfrak{q}}\mathrm{\ne }0$ by step 1.1 and $J{A}_{\mathfrak{q}}$ lies in the maximal ideal of the local ring $A_{\mathfrak{q}}$; [L6] applied there gives $Q_{\mathfrak{q}}\mathrm{\ne }0$. Hence ${\mathrm{Supp}}_A(Q)=V_A(JA)$. By [L5], this support is exactly $\{\mathfrak{n}\}$ if and only if $\sqrt{JA}=\mathfrak{n}$.

Combining steps 2.1 and 1.2, for ideals $J\subseteq \mathfrak{m}$ the quotient $M\mathrm{/}JM$ has finite length exactly when $\sqrt{JA}=\mathfrak{n}$. Applying [L1] in $A$, the least size of a generating tuple in $\mathfrak{n}$ with this property is $d$. Every such tuple in $A$ lifts to a tuple in $\mathfrak{m}$ because the quotient map $R\to A$ sends $\mathfrak{m}$ onto $\mathfrak{n}$; finite choices suffice. This proves part 1.

If $x_1,\dots ,x_d\in \mathfrak{m}$ satisfy that $M\mathrm{/}(x_1,\dots ,x_d)M$ has finite length, step 3.1 gives $\sqrt{(x_1,\dots ,x_d)A}=\mathfrak{n}$. Their images in $A$ therefore form a system of parameters by [L2]; this is the asserted module parameter condition. For $d=0$, the tuple is empty and the same argument says $M$ itself has finite length.

Steps 3.1 and 4.1 prove both claims.