Rees Modules Artin Rees and Hilbert Samuel Theory

If $r=0$, then $S=A$ and the finite graded $A$-module $M$ has only finitely many nonzero homogeneous pieces. Hence ${\mathrm{HS}}_M(t)$ is a Laurent polynomial, so both conclusions hold.

Assume $r>0$ and 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 $0\to K\to M(-1)\stackrel{x_r}{\to }M\to C\to 0,$ where $K$ and $C$ are annihilated by $x_r$ and therefore are finite graded $S^{\mathrm{\prime }}$-modules.

Taking degree-$n$ pieces in algebra 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].

By the algebra hypothesis applied to the finite graded $S^{\mathrm{\prime }}$-modules $K$ and $C$, the two series on the right side of algebra have denominator dividing $(1-t{)}^{r-1}$. Therefore ${\mathrm{HS}}_M(t)$ has denominator dividing $(1-t{)}^r$.

Any rational function with denominator a power of $(1-t)$ expands for large $n$ as a finite $\mathbb{Z}$-linear combination of binomial coefficients $\left(\genfrac{}{}{0px}{}{n+d}{d}\right)$, hence its coefficients agree eventually with a polynomial. Applying this to algebra proves the eventual polynomial behaviour of $H_M(n)={\mathrm{\ell }}_A(M_n)$.

Steps 1.1 through 1.5 prove both claims.

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.

Because $M\mathrm{/}IM$ has finite length over the local ring $(R,\mathfrak{m})$, some power ${\mathfrak{m}}^c$ annihilates it. Equivalently, ${\mathfrak{m}}^{c}M\subseteq IM$. Multiplying by $I^n$ gives ${\mathfrak{m}}^c{I}^{n}M\subseteq I^{n+1}M$ for every $n\ge 0$, so each graded piece $I^{n}M\mathrm{/}{I}^{n+1}M$ is naturally a module over the Artinian local ring $A:=R\mathrm{/}{\mathfrak{m}}^c$.

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: ${\chi }_{I,M}(n)={\sum }_{j=0}^n{\phi }_{I,M}(j)\mathrm{.}$ Summing a polynomial tail again produces a polynomial tail, and summing the standard binomial basis $\left(\genfrac{}{}{0px}{}{j+d-1}{d-1}\right)$ produces $\left(\genfrac{}{}{0px}{}{n+d}{d}\right)$. Therefore ${\chi }_{I,M}(n)$ is eventually a polynomial in 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 $A:=R\mathrm{/}{\mathrm{Ann}}_R(M)$. Then $M$ is a faithful finite $A$-module and ${\mathrm{Supp}}_R(M)$ identifies with $\mathrm{Spec}(A)$, so $d=\dim A$. The maximal ideal of the Noetherian local ring $A$ has finitely many generators by [L3], and [L4] bounds its height by that finite number. Thus $d<\mathrm{\infty }$, so [L1] applies to $A$.

For an ideal $J\subseteq R$, the quotient $M\mathrm{/}JM$ has finite length if and only if $(A\mathrm{/}JA{)}_{\mathfrak{p}}=0$ for every nonmaximal prime $\mathfrak{p}$ of $A$, which is equivalent to $\sqrt{JA}={\mathfrak{m}}_A$. Thus ideals of definition for $M$ are exactly the ideals of $A$ with maximal radical.

Applying [L1] in the local ring $A$, the least number of generators of an ideal with radical ${\mathfrak{m}}_A$ is exactly $\dim A=d$. By algebra this is the least number of generators of an ideal of definition for $M$.

If $x_1,\dots ,x_d\in \mathfrak{m}$ satisfy that $M\mathrm{/}(x_1,\dots ,x_d)M$ has finite length, then algebra gives $\sqrt{(x_1,\dots ,x_d)A}={\mathfrak{m}}_A\mathrm{.}$ Therefore their images in $A$ form a system of parameters by [L2], and we call the original tuple a system of parameters for $M$.

This proves both claims.

Artin-Rees as recorded in [L1] gives $c\ge 0$ and $N:=M^{\mathrm{\prime }}\cap I^{c}M$ such that $I^c{M}^{\mathrm{\prime }}\subseteq N\subseteq M^{\mathrm{\prime }}$ and $M^{\mathrm{\prime }}\cap I^{n+1}M=I^{n+1-c}N$ for all large $n$. The exact sequence $0\to M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n+1}M)\to M\mathrm{/}{I}^{n+1}M\to M^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{I}^{n+1}{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ and [L2] therefore give ${\chi }_{I,M}(n)={\chi }_{I,M^{\mathrm{\prime }\mathrm{\prime }}}(n)+{\chi }_{I,N}(n-c)+{\mathrm{\ell }}_R(M^{\mathrm{\prime }}\mathrm{/}N)$ for all large $n$.

Put $C:={\mathrm{\ell }}_R(M^{\mathrm{\prime }}\mathrm{/}N)$. The inclusions $I^c{M}^{\mathrm{\prime }}\subseteq N\subseteq M^{\mathrm{\prime }}$ give $I^{n+c+1}{M}^{\mathrm{\prime }}\subseteq I^{n+1}N\subseteq I^{n+1}{M}^{\mathrm{\prime }},$ while [L2] gives ${\mathrm{\ell }}_R(M^{\mathrm{\prime }}\mathrm{/}{I}^{n+1}N)={\chi }_{I,N}(n)+C\mathrm{.}$ Consequently ${\chi }_{I,M^{\mathrm{\prime }}}(n)\le {\chi }_{I,N}(n)+C\le {\chi }_{I,M^{\mathrm{\prime }}}(n+c)$ for all large $n$. If $M^{\mathrm{\prime }}$ has positive Hilbert-Samuel degree, this squeeze shows that $P_{I,N}$ and $P_{I,M^{\mathrm{\prime }}}$ have the same degree and leading coefficient. If $P_{I,M^{\mathrm{\prime }}}$ has degree zero, the two outer terms in the squeeze are the same constant polynomial, so $P_{I,N}+C=P_{I,M^{\mathrm{\prime }}}$. The same conclusion is immediate when $M^{\mathrm{\prime }}=0$.

The eventual identity in step 1.1 is the polynomial identity $P_{I,M}(n)=P_{I,M^{\mathrm{\prime }\mathrm{\prime }}}(n)+P_{I,N}(n-c)+C\mathrm{.}$ By step 2.1, the polynomial $P_{I,N}(n-c)+C$ has the same degree-$d$ coefficient as $P_{I,M^{\mathrm{\prime }}}$: for positive degree this is invariance of the leading coefficient under a shift, for degree zero it is the constant-polynomial equality, and below degree $d$ both coefficients vanish. Comparing degree-$d$ coefficients therefore gives $e_I^{[d]}(M)=e_I^{[d]}(M^{\mathrm{\prime }})+e_I^{[d]}(M^{\mathrm{\prime }\mathrm{\prime }})\mathrm{.}$ This also covers the all-zero sequence by the convention $d=0$. When all three modules are nonzero of degree $d$, the displayed quantities are their ordinary Hilbert-Samuel multiplicities.

Therefore Hilbert-Samuel leading coefficients are additive in the stated top-degree sense.