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Whenever the graded pieces $N_i$ are finitely generated over $S_0$, define the Poincare series of a graded module byAdditivity means $\lambda (B)=\lambda (A)+\lambda (C)$ for every short exact sequence $0\to A\to B\to C\to 0$ in the indicated category. Dimension over a field is a basic example; length is another when the modules have finite length.

The Hilbert-Serre theorem needs finite-generation hypotheses: take $S_0$ commutative Noetherian and $S$ generated over it by homogeneous $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$. For finitely generated graded $N$, every $N_i$ is then finite over $S_0$, andIf negative grading indices are allowed for the module, the numerator is a Laurent polynomial instead. In standard degree one, the coefficients are eventually polynomial functions of the index.

To prove this, regard $N$ as finite over the surjective polynomial presentation $S_0[X_1,\dots ,X_r]\to S$. Induct on $r$. At $r=0$ only finitely many homogeneous generator degrees occur, so the series is a polynomial. For the last variable $x$ of degree $d$, put $K=(0{:}_{N}x)$ and $Q=N/xN$. The polynomial ring is Noetherian by the Hilbert basis theorem, hence $K$ is finite; both $K$ and $Q$ are finite modules over the ring with that variable removed. The degree-$i$ exact sequence isAdditivity, followed by summing over $i$, givesThe induction hypothesis supplies the remaining denominator factors. This proves Hilbert-Serre theorem for an additive coefficient function. If all $d_i=1$, expanding $(1-t{)}^{-r}$ proves eventual polynomial coefficients directly.

An arbitrary positively graded ring in the introductory wording is not enough by itself. For example $S=k[X_1,X_2,\dots ]$ with every variable of degree one and $N=S$ is a cyclic graded module, but $N_1$ is not finite-dimensional over $k$, so the dimension-valued series is not even defined. The theorem above states the standard hypotheses that make the requested definition and rationality valid.

A precise version of the Hilbert-Serre theorem is as follows. Let $R={⨁}_{j\ge 0}{R}_j$ be a commutative graded ring, generated over an Artinian degree-zero ring $R_0$ by homogeneous elements $y_1,\dots ,y_s$ of positive degrees $d_1,\dots ,d_s$. Let $M$ be a finitely generated graded module, bounded below. Each component has finite module length over $R_0$. Its Poincare series of a graded module, or Hilbert series, isWhen $R_0$ is a field, these lengths are dimensions. The theorem assertsFor a nonnegatively graded module the numerator can be taken in $\mathbb{Z}[t]$. The finite-generation and finite-length hypotheses specify the meaning of the series; they are part of the theorem's statement.

We prove it by induction on $s$. Regard $M$ as a finite graded module over $S=R_0[Y_1,\dots ,Y_s]$, with $\mathrm{deg}{Y}_i=d_i$, acting through the specified generators of $R$. This polynomial ring is Noetherian by the Hilbert basis theorem and Question 1. When $s=0$, finitely many homogeneous generators over $R_0$ mean that only finitely many components are nonzero, so the series is a Laurent polynomial. For the induction step, put $y=Y_s$, $d=d_s$, $K=(0{:}_{M}y)$ and $C=M/yM$. Both $K$ and $C$ are finite graded modules; $y$ annihilates them, so they are modules over $R_0[Y_1,\dots ,Y_{s-1}]$. The finite generation of $K$ uses Noetherianity of $S$.

The exact sequence of graded modulesgives, by additivity of component lengths,Here the shift is defined so that $P_{M(-d)}=t^d{P}_M$. By induction both series on the right have denominator ${\prod }_{i<s}(1-t^{d_i})$. Dividing by $1-t^d$ proves the result. The annihilator term cannot be omitted unless multiplication by $y$ is injective.

In the standard grading $d_i=1$, the coefficients of $(1-t{)}^{-s}$ are $\left(\genfrac{}{}{0ex}{}{j+s-1}{s-1}\right)$. Multiplying by the finite numerator shows that the component dimensions agree with a polynomial for sufficiently large $j$. Cumulative dimensions have generating function $P_M(t)/(1-t)$ and are also eventually polynomial. This consequence supplies the dimension used for Weyl algebra modules.

Use the Bernstein filtration $B_m{A}_n$, in which both coordinate generators and differentiation generators have degree one. The ordered monomial basis of a Weyl algebra givesChoose a finite-dimensional generating subspace $M_0$ of a nonzero finitely generated left module $M$ and set $M_m=B_m{A}_n{M}_0$. This is a good filtration of a module: the symbols of the generators generate its associated graded module over the displayed polynomial ring. The Hilbert-Serre theorem makes ${\dim }_{\mathbb{C}}{M}_m$ an eventual polynomial. Define the Bernstein growth dimension of a Weyl algebra module byThe dimension is independent of the chosen finite generating subspace. Each of two finite generating sets is contained in a fixed filtration step for the other, giving $M_m\subseteq M_{m+c}^{\prime }$ and $M_m^{\prime }\subseteq M_{m+c^{\prime }}$. Such fixed shifts preserve the degree of growth. The same argument applies to any good Bernstein filtration. This dimension agrees with the Gelfand–Kirillov dimension of a module; one may set $d(0)=-\infty$.

Let the nonconstant element $D$ have Bernstein degree $d\ge 1$. Since ${\mathrm{gr}}_B{A}_n$ is a domain, the product of two nonzero symbols cannot vanish. HenceRight multiplication by $D$ is injective, using the domain property of $A_n$. Thus the quotient filtration on $M=A_n/A_{n}D$ has cumulative dimensionsThe leading terms of degree $2n$ cancel, and the leading surviving term isIt is nonzero, because $d\ge 1$. ThereforeEquivalently the Hilbert series of the associated graded quotient is $(1-t^d)/(1-t{)}^{2n}$. Here $d$ is Bernstein degree, not differential order: multiplication by $x_1$ has differential order zero but Bernstein degree one. Using the order filtration in this dimension argument would lose the finite-dimensional filtration pieces.

Use the following finite-length version of the Hilbert-Serre theorem. Let $S={⨁}_{n\ge 0}{S}_n$ be a graded algebra generated over an Artinian ring $A=S_0$ by finitely many homogeneous elements $x_1,\dots ,x_r$ of positive degrees $e_1,\dots ,e_r$. For a finitely generated nonnegatively graded $S$-module $M$, define its Poincare series of a graded module byEvery component has finite length, and the theorem saysFor a module whose grading is merely bounded below, the same statement holds with a Laurent-polynomial numerator. When the generators all have degree one, the denominator is $(1-t{)}^r$.

Here is an induction proof. An Artinian ring is Noetherian by question 1, so the Hilbert basis theorem makes $S$ Noetherian. If there are no positive-degree generators, $S=A$ and a finite homogeneous generating set for $M$ occupies only finitely many degrees. Each component is a finite module over the Artinian ring $A$, hence has finite length of a module, and $P_M$ is a polynomial.

For $r>0$, put $x=x_r$, $e=e_r$, $N=\{m\in M:xm=0\}$ and $C=M/xM$. Both are finitely generated graded modules annihilated by $x$, hence modules over $S/(x)$, which is generated by the first $r-1$ homogeneous elements. Their multiplication exact sequence isHere $M(-e{)}_n=M_{n-e}$. Additivity of length degree by degree givesThis is the Hilbert series multiplication exact sequence. By induction, the right-hand side has denominator ${\prod }_{i<r}(1-t^{e_i})$. Division by $1-t^e$ completes the proof of the Hilbert-Serre theorem. The finite-component and finite-generation claims also follow from the finite set of positive-degree algebra and module generators; no analytic convergence of a series is involved.

For the local invariant, use the usual Noetherian local ring hypothesis of Hilbert–Samuel growth dimension. Locality alone does not guarantee finite lengths or polynomial growth; the printed question leaves this finiteness assumption implicit. For instance, in the localization at a prime ideal of the polynomial ring $k[x_1,x_2,\dots ]$ at $(x_1,x_2,\dots )$, the vector space $\mathfrak{m}/{\mathfrak{m}}^2$ has the infinitely many independent classes of the variables, so its length is not finite. Write $\mathfrak{m}=P$ and $\kappa =R/\mathfrak{m}$. Its associated graded ringis generated over $\kappa$ in degree one, because $\mathfrak{m}$ is finitely generated. Thus its Hilbert series is rational with a denominator that is a power of $1-t$. After cancelling factors, writeDefine $d(R)=d$, the pole order at $t=1$, with $d=0$ when $P_G$ is a polynomial.

Equivalently, the Hilbert–Samuel functionhas generating series $P_G(t)/(1-t)$ and eventually agrees with a polynomial of degree $d$. Indeed, coefficients of $(1-t{)}^{-d-1}$ are $\left(\genfrac{}{}{0ex}{}{n+d}{d}\right)$, and multiplication by $q(t)$ leaves leading term $q(1)n^d/d!$. Consequently $d(R)$ is the degree of the cumulative Hilbert-Samuel polynomial, rather than the degree of the individual graded-component function; the latter has degree $d-1$ when $d>0$. This pole/growth invariant also equals Krull dimension by the local dimension theorem, although that theorem is not required to define it here.

For an example, take $R=k[x,y{]}_{(x,y)}$. Its associated graded ring is $k[x,y]$ with the ordinary degree grading: degree $n$ has the $n+1$ monomials $x^i{y}^{n-i}$ as a basis. ThereforeThe length formula also follows directly by counting monomials of total degree at most $n$ in $R/(x,y{)}^{n+1}$.

For a finitely generated module $V={⨁}_d{V}_d$ over the graded ring $S=k[X_1,\dots ,X_n]$, with each $X_i$ of degree one, its Poincare series of a graded module isThis is also its Hilbert series. One usually takes $d\ge 0$; allowing an integer grading also permits finitely many negative degrees. Homogeneous generators of degrees $a_1,\dots ,a_h$ give a surjection ${⨁}_{j}S(-a_j)\to V$, where the graded shift is defined by $V(-a{)}_d=V_{d-a}$. Consequently every $V_d$ is a finite-dimensional vector space and $V_d=0$ for all sufficiently negative $d$.

The Hilbert-Serre theorem in the standard grading saysFor a nonnegatively graded $V$, the numerator is in $\mathbb{Z}[t]$. Equivalently, the only possible pole of this rational expression is at $t=1$, with order at most $n$.

We prove the Hilbert-Serre theorem by induction on the number of variables. For $n=0$, $S=k$ and $V$ is a finite-dimensional graded module; thus $P_V(t)$ is a Laurent polynomial. Suppose $n>0$ and putThe Hilbert basis theorem makes $S$ a Noetherian ring, so $K$ is a finitely generated module, as is $C$. Both are graded modules killed by $X_n$, hence finitely generated over $S^{\prime }=k[X_1,\dots ,X_{n-1}]$.

The graded exact sequenceidentifies the kernel and cokernel of multiplication by $X_n$. Taking the alternating sum of the finite-dimensional degree components yieldssoBy induction the right side has denominator $(1-t{)}^{n-1}$, which proves the required denominator $(1-t{)}^n$. Since $(1-t{)}^n{P}_V(t)$ has no negative powers when $V$ is nonnegatively graded, its Laurent polynomial numerator is an ordinary polynomial in that case.

For $n\ge 1$, this also implies the usual Hilbert polynomial consequence. Writing $Q(t)={\sum }_j{q}_j{t}^j$ and usinggives, for all sufficiently large $d$,a polynomial in $d$ of degree at most $n-1$. For $n=0$, the graded pieces are eventually zero.

For the final request, a free resolution of $V$ is an exact sequencewith each $F_i$ a free module. In the graded setting we take finite direct sums of shifts $S(-a)$ and maps preserving degree. The resolution has length at most $n$ if $F_i=0$ for every $i>n$. The syzygy modules are the successive kernels which record the relations among generators, then relations among those relations, and so on.

Put $\mathfrak{m}=(X_1,\dots ,X_n)$, the homogeneous maximal ideal with $S/\mathfrak{m}=k$. Choose a homogeneous $k$-basis of $V/\mathfrak{m}V$ and lift it to $V$. These lifts generate $V$ by the graded Nakayama lemma: a bounded-below graded module $W$ satisfying $W=\mathfrak{m}W$ must be zero, since a nonzero homogeneous element of least degree could not be a sum of variables times elements of lower degrees. Apply this to the quotient by the submodule generated by the chosen lifts. This gives a surjection $F_0\to V$ inducing an isomorphism modulo $\mathfrak{m}$.

Its kernel is finitely generated because $S$ is a Noetherian ring. Repeat the construction for that kernel, then for each subsequent kernel. We obtain a minimal graded free resolution, meaningIndeed, at each stage the free cover induces an isomorphism modulo $\mathfrak{m}$, so its kernel lies in $\mathfrak{m}$ times its source. The sequence is exact by construction, though it may at first appear infinite.

The Koszul complex on $X_1,\dots ,X_n$ iswith basis vectors $e_j$ given degree one, and differentialThe hat means omission of that factor. Every pair of terms in ${\partial }^2$ cancels with opposite signs, so this is a chain complex. The sequence $X_1,\dots ,X_n$ is a regular sequence: after quotienting by the first $j-1$ variables, the next variable is a non-zero-divisor in the remaining polynomial ring.

The Koszul complex is consequently exact in positive degrees and has $H_0=k$, so it is a Koszul resolution of $k$ of length $n$. For completeness, one proves this by induction: for one variable it is $0\to S\stackrel{X_1}{\to }S\to S/(X_1)\to 0$. Appending $X_j$ constructs the mapping cone of multiplication by $X_j$ on the previous chain complex. Its homology is zero in positive degrees because $X_j$ acts injectively on $S/(X_1,\dots ,X_{j-1})$; its degree-zero homology is the further quotient. Here a mapping cone combines a chain complex with a shifted copy and adds the given multiplication map to the differential, thereby measuring its kernel and cokernel on homology.

The Tor functor ${\mathrm{Tor}}_i^S(V,k)$ is defined as the degree-$i$ homology of $F_{\bullet }{\otimes }_{S}k$ for a free resolution $F_{\bullet }$ of $V$. It can equally be computed as the homology of $V{\otimes }_S{K}_{\bullet }$, using the Koszul resolution of $k$. To see the equality, form the double complex $F_{\bullet }{\otimes }_S{K}_{\bullet }$. Taking homology first along $F_{\bullet }$ leaves $V{\otimes }_S{K}_{\bullet }$, since each $K_i$ is a free module; taking homology first along $K_{\bullet }$ leaves $F_{\bullet }{\otimes }_{S}k$, since each $F_i$ is free. The finite sums along each total degree identify both with the homology of the total complex. This also explains the symmetry used in computing the Tor functor.

Since $K_i=0$ for $i>n$, we have ${\mathrm{Tor}}_i^S(V,k)=0$ for $i>n$. On the other hand, all differentials of the minimal graded free resolution become zero after tensoring with $k=S/\mathfrak{m}$, soThus $F_i/\mathfrak{m}{F}_i=0$ for $i>n$, and the graded Nakayama lemma implies $F_i=0$ for those $i$. We have proved the required Hilbert syzygy theorem:where some initial $F_i$ may be zero. The argument supplies finitely generated graded free modules, which is stronger than merely giving an ungraded free resolution.

Choose homogeneous algebra generators $x_1,\dots ,x_r$ of positive degrees $d_1,\dots ,d_r$ for $S$; they exist because $S$ is Noetherian. For a finitely generated graded module $V={⨁}_j{V}_j$, its Poincare series of a graded module isThe Hilbert-Serre theorem states thatfor a Laurent polynomial $Q(t)\in \mathbb{Z}[t,t^{-1}]$.

We prove this by induction on $r$. For $r=0$, $S=k$ and $V$ is finite-dimensional, so its series is a Laurent polynomial. For $x=x_r$ of degree $d=d_r$, multiplication gives the exact sequence of graded modulesAdditivity of the Hilbert series yieldsBoth modules on the right are finitely generated over $k[x_1,\dots ,x_{r-1}]$, so the induction hypothesis proves the formula.

For the finitely generated commutative algebra $R$, let $R_j$ be the stated degree filtration. Its associated graded ringis a standard graded algebra generated by the initial forms of $x_1,\dots ,x_n$. SinceHilbert–Serre shows that the quantity ${\dim }_k{R}_j$ agrees with a polynomial for all sufficiently large $j$. The growth of a finitely generated commutative algebra has polynomial degree $\dim R$, independently of the chosen finite generating set.

Forwith the four displayed generators, a basis of $R_j$ consists of the Laurent monomials $Y_1^a{Y}_2^b$ satisfying $|a|+|b|\le j$. There are $4r$ points with $|a|+|b|=r>0$, and one point at the origin. Hence the growth of the two-variable Laurent polynomial algebra is

For an additive length function $\lambda$ finite on the graded pieces, define the Poincare series of a graded moduleThe Hilbert-Serre theorem states thatfor a Laurent polynomial $f(t)\in \mathbb{Z}[t,t^{-1}]$.

Induct on $s$. For the last generator $x_s$ of degree $k_s$, multiplication gives an exact sequence whose kernel is the $x_s$-torsion and whose cokernel is $M/x_{s}M$. Additivity of $\lambda$ yieldsBoth modules on the right are finite graded modules over the algebra generated by $x_1,\dots ,x_{s-1}$. The induction hypothesis gives the asserted denominator. The case $s=0$ is a finite Laurent polynomial because $M$ is finitely generated over $A_0$.

The Poincare series of a graded module isIt is the generating function of the Hilbert function $n\mapsto {\dim }_k{A}_n$.

The weighted Hilbert series of $K[x,y,t_1,\dots ,t_n]$ isThe homogeneous polynomial $x^k-y^k$ has degree $k$ and is a non-zero-divisor, so quotienting by it multiplies the series by $1-z^k$. Therefore the requested Poincare series of a graded module is