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For a good filtration of a module over a fixed finite-dimensional filtration of a ring whose associated graded algebra is commutative and finitely generated, the cumulative Hilbert function is eventually a quasipolynomial by the Hilbert-Serre theorem. Its leading asymptotic degree is the dimension of the nonzero filtered module. Monotonicity forces the leading coefficients in its residue classes to agree. If the associated graded algebra is a standard graded algebra, the Hilbert function is eventually polynomial. Two good module filtrations bound one another after fixed shifts, so they have the same growth degree. Set $d(0)=-\infty$. For the Bernstein filtration this is the Bernstein growth dimension of a Weyl algebra module.

Let a commutative $k$-algebra $R$ be generated by $x_1,\dots ,x_n$, and let $R_j$ be the span of words of total degree at most $j$ in those generators. The associated graded ring of this filtration is a standard graded algebra, andThe Hilbert-Serre theorem therefore makes ${\dim }_k{R}_j$ eventually a polynomial in $j$. Its degree is $\dim R$; this follows by applying the Rees-ring deformation, whose special fibers are $R$ and the associated graded ring, so they have the same Krull dimension.

For a homogeneous element $x$ of degree $e>0$ acting on a finite graded module, set $N=(0{:}_{M}x)$ and $C=M/xM$. The multiplication exact sequence $0\to N(-e)\to M(-e)\to M\to C\to 0$ and additivity of component lengths give $(1-t^e)P_M=P_C-t^e{P}_N$. This is the inductive step in the Hilbert-Serre theorem; the annihilator term must be kept unless multiplication by $x$ is injective.

Work over $\mathbb{C}$, or another field of characteristic zero. The Weyl algebra $A_n$ is generated by $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ with commuting $x$'s, commuting $\partial$'s and $[{\partial }_i,x_j]={\delta }_{ij}$. Its ordered monomial basis of a Weyl algebra consists of the monomials $x^{\alpha }{\partial }^{\beta }$. One way to verify independence is to apply a finite operator relation to the formal exponential $e^{\mathbf{t}\cdot \mathbf{x}}$: after canceling that exponential the relation becomes a polynomial identity $\sum c_{\alpha \beta }{\mathbf{x}}^{\alpha }{\mathbf{t}}^{\beta }=0$, so every coefficient vanishes. The relations already reorder every word into that basis.

Give both generators degree one. The Bernstein filtration isThus $\dim F_m{A}_n=\left(\genfrac{}{}{0ex}{}{m+2n}{2n}\right)$. Commutator terms have lower degree, accounting for the commutative symbol algebra. Since that algebra is Noetherian, leading-symbol generation followed by finite downward degree cancellation proves that $A_n$ is Noetherian on both sides.

A left finitely generated module $M$ has a good filtration, for example $M_m={\sum }_j{F}_m{A}_n{m}_j$ for a finite generating set of the module. More generally one allows fixed degree shifts of the generators. The associated graded module is finite over the symbol algebra. DefineThis is the characteristic variety of a Weyl algebra module for the Bernstein algebra filtration. It records where the module's leading symbols remain nonzero and makes noncommutative module growth accessible to commutative algebra.

For a fixed algebra filtration it is independent of the good filtration. Any two such filtrations differ by bounded shifts $M_i\subseteq {\tilde{M}}_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\mathrm{gr}M$, its $r$th power maps $M_i$ into $M_{i+rd-r}$. For $r>2c$, bounded-shift comparison forces the power of that symbol to annihilate $\mathrm{gr}\tilde{M}$. Reversing the argument gives equal radicals of ideals of annihilators. This is independence of characteristic support from a good module filtration. Changing the algebra filtration itself, for example to the order filtration of a Weyl algebra with the $x_i$ in degree zero, need not preserve the same variety; the convention matters.

The Hilbert-Serre theorem gives an eventual cumulative polynomialHere $d$ is also the Gelfand–Kirillov dimension of a module, and $e(M)>0$ is an integer multiplicity of a filtered module for nonzero $M$. Finite generation bounds $d\le 2n$.

Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_m{A}_n$, choose a nonzero monomial of greatest total degree in its ordered expansion, say $x^{\alpha }{\partial }^{\beta }$. The operations $\mathrm{ad}{\partial }_i$ and $\mathrm{ad}(-x_i)$ differentiate this expansion with respect to $x_i$ and ${\partial }_i$. Apply them ${\alpha }_i$ and ${\beta }_i$ times. Lower-degree terms die, and every other monomial of the same degree also dies unless it has exactly those exponents. The result is a nonzero scalar multiple of the chosen coefficient. Characteristic zero is essential for the factorials. Expanding the repeated commutators therefore givesChoose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $f{v}_{j}v$ would vanish, contrary to $v={\sum }_j{u}_{j}f{v}_{j}v$. Hence the linear action mapis injective. It follows that $\dim F_m{A}_n\le (\dim M_m)(\dim M_{2m})$. Comparing polynomial degrees gives $2n\le 2d$. ThusThis proves Bernstein inequality for Weyl algebra modules through the faithful finite-step action of a Weyl algebra.

The bound is sharp. The polynomial representation $\mathbb{C}[x_1,\dots ,x_n]\cong A_n/{\sum }_i{A}_n{\partial }_i$ has the zero section as its characteristic variety of a Weyl algebra module, namely $\xi =0$ and dimension $n$. It is a simple module: differentiate a nonzero polynomial sufficiently to obtain a nonzero constant, then multiply by the $x_i$ to generate the whole module. The module $A_n/{\sum }_i{A}_n{x}_i$ similarly has characteristic variety of a Weyl algebra module $x=0$, a fiber of the cotangent bundle. A nonzero free module has the whole cotangent bundle and dimension $2n$. No nonzero finite-dimensional module exists for $n\ge 1$, also seen by taking the trace of $[{\partial }_1,x_1]=I$.

Modules attaining dimension $n$ are holonomic Weyl algebra modules (with the zero module included conventionally). Every nonzero subquotient remains of dimension $n$ by the lower bound and monotonicity of growth. Compatible filtrations in a short exact sequence make the leading multiplicity of a filtered module additive. Since each nonzero factor has positive integer multiplicity of a filtered module, the number of factors in any strict submodule chain is bounded by $e(M)$. Together with Noetherianity, this proves finite composition length. It illustrates why characteristic variety of a Weyl algebra module and multiplicity of a filtered module provide useful structure beyond a list of presentations. In positive characteristic there are finite-dimensional Weyl modules, so the characteristic-zero inequality and this discussion must not be asserted without its field hypothesis.

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.

Fix a finite-dimensional nonnegative filtration of a ring giving $R$ its almost commutative algebra structure, and choose a good filtration of a module on $M$. When $\mathrm{gr}R$ is a standard graded algebra, the Hilbert-Serre theorem makes the cumulative Hilbert function $h_M(j)={\dim }_k{F}_{j}M$ eventually a polynomial. For $M\ne 0$, define the dimension of a filtered module and multiplicity of a filtered module byUse $d(0)=-\infty$ and $m(0)=0$. With positive weighted generators instead of a standard grading, the cumulative Hilbert function is eventually a quasipolynomial with a common positive leading coefficient across residue classes. Equivalently, $h_M(j)=c{j}^d+O(j^{d-1})$, and the definitions are $d(M)=d$ and $m(M)=d!c$. The common leading coefficient follows from monotonicity: interlacing consecutive residue classes forces their leading coefficients to agree. For a standard grading, multiplicity is a positive integer, since an integer-valued polynomial has an integral expansion in the binomial polynomials; weighted normalization can instead give a positive rational number.

Two good filtrations of a module for this fixed algebra filtration bound each other after fixed index shifts. The corresponding Hilbert functions therefore satisfy $h_F(j-a)\le h_G(j)\le h_F(j+b)$ for fixed $a,b$. Comparing their leading growth shows that the degree and leading coefficient agree. Thus $d(M)$ and $m(M)$ are independent of the module filtration; the fixed algebra filtration is part of the multiplicity convention.

For a submodule $N$, use its subspace filtration and the quotient filtration on $M/N$. These are good because $\mathrm{gr}R$ is Noetherian. The short exact sequence of filtered pieces givesNonzero leading coefficients are positive, soIf $d(N)=d(M/N)$, those leading coefficients add andThis is dimension and multiplicity in a filtered exact sequence.

For the Weyl algebra $A_n(k)$ in characteristic zero, Bernstein inequality for Weyl algebra modules asserts $d(M)\ge n$ for every nonzero finitely generated module $M$. Write its generators as $x_i,D_i$, with $[D_i,x_j]={\delta }_{ij}$ and all other generator commutators zero. The ordered monomials $x^{\alpha }{D}^{\beta }$ give the Bernstein filtrationTake a finite-dimensional generating subspace $V\ne 0$ of $M$ and put $M_j=A_{j}V$.

To prove faithful finite-step action of a Weyl algebra, consider $0\ne a\in A_j$ and choose a nonzero coefficient $c_{\alpha \beta }$ of a monomial of maximal total degree $d\le j$. Applying ${\prod }_i(\mathrm{ad}{D}_i{)}^{{\alpha }_i}(\mathrm{ad}{x}_i{)}^{{\beta }_i}$ to $a$ givesIndeed, these commutators differentiate the ordered coefficients: lower-degree monomials vanish, and another degree-$d$ monomial can survive only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. This is scalar extraction by iterated Weyl commutators.

Expanding these iterated commutators writes $\lambda ={\sum }_{\ell }{u}_{\ell }a{v}_{\ell }$ with $u_{\ell },v_{\ell }\in A_j$. If $a$ annihilated $M_j$, every summand would annihilate $V$, contrary to $\lambda V\ne 0$. Therefore the action map is injective:Taking dimensions and comparing polynomial degrees giveswhich proves Bernstein inequality for Weyl algebra modules. The polynomial representation $k[x_1,\dots ,x_n]$ has dimension $n$, so the bound is sharp.

The associated graded ring of the Weyl algebra is a polynomial ring, so ascending filtered-graded transfer of Noetherianity makes $A_n(k)$ a left Noetherian ring. Thus every submodule of a finitely generated module is finitely generated. If $d(M)=n$, every nonzero subquotient has dimension at most $n$ by the exact-sequence formula, and at least $n$ by Bernstein inequality for Weyl algebra modules. Every such subquotient therefore has dimension $n$ and positive integral multiplicity.

In a strictly descending chain $M=M_0⊋M_1⊋\cdots$, each nonzero quotient $M_i/M_{i+1}$ consumes at least one unit of multiplicity, by dimension and multiplicity in a filtered exact sequence. Hence there are at most $m(M)$ strict steps. This proves the descending chain condition, so $M$ is Artinian. In fact this holonomic Weyl algebra module has module length at most $m(M)$, as in multiplicity bounds the length of a holonomic Weyl module.

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 $R/(x,y{)}^{n+1}$.

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.

For $a\in R$, write $m_a$ for multiplication by $a$ in ${\mathrm{End}}_k(R)$. The algebraic differential operator ring is defined by the filtrationThe order of an algebraic differential operator is the least $j$ for which $\theta \in D_k^j(R)$; the zero operator may be assigned order $-\infty$. In degree zero, commuting with all multiplication maps means $\theta (a)=a\theta (1)$, so $D_k^0(R)=\{m_a:a\in R\}\cong R$. The commutator identity $[UV,m_a]=U[V,m_a]+[U,m_a]V$ proves by induction that $D_k^r{D}_k^s\subseteq D_k^{r+s}$, so the union is a ring under addition and composition.

The derivations of an algebra formEvery such derivation has $\delta (1)=0$ and $[\delta ,m_a]=m_{\delta (a)}$, so has differential order at most one. Conversely, if $\theta \in D_k^1(R)$, set $c=\theta (1)$ and $\delta =\theta -m_c$. Then $\delta (1)=0$ and $[\delta ,m_a]$ is a multiplication map. Evaluating at $1$ shows it is $m_{\delta (a)}$; evaluating at $b$ gives the Leibniz rule. Hence the first-order algebraic differential operator decomposition isIt is a direct sum as $k$-vector spaces: a multiplication map that is a derivation vanishes at $1$, and is zero.

The complex Weyl algebra has generators $x_1,\dots ,x_n,{\partial }_1,\dots ,{\partial }_n$ and relationsTake $n\ge 1$. If a nonzero unital module $M$ had finite dimension of a vector space over $\mathbb{C}$ $m$, its representing matrices would satisfy $[{\partial }_1,x_1]=I_m$. Taking the matrix trace gives $0=m$, impossible in characteristic zero. The Weyl algebra has no nonzero finite-dimensional modules. For $n=0$ the algebra is $\mathbb{C}$ and this assertion has the obvious exception.

For a nonzero finitely generated Weyl algebra module, use the Bernstein filtrationThe ordered monomials form a vector space basis, soTo justify the ordered basis, the relations move all ${\partial }_i$ to the right and give spanning. Independence follows in the action on $\mathbb{C}[x_1,\dots ,x_n]$: apply a putative zero operator ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x){\partial }^{\mathbf{b}}$ coefficientwise to the formal exponential $e^{t\cdot x}$, obtaining ${\sum }_{\mathbf{b}}{p}_{\mathbf{b}}(x)t^{\mathbf{b}}=0$. Thus every coefficient vanishes.

Choose a finite-dimensional nonzero generating subspace $M_0\subseteq M$ and set $M_j=F_j{A}_n{M}_0$. The associated graded module is finitely generated over the polynomial associated graded ring, so the Hilbert-Serre theorem implies ${\dim }_{\mathbb{C}}{M}_j$ is eventually a polynomial. DefineThis Bernstein growth dimension of a Weyl algebra module equals the Gelfand–Kirillov dimension of a module, not its ordinary vector-space dimension. If a different finite generating subspace is used, each lies in some fixed filtration step of the other. The corresponding $M_j$ are sandwiched between shifted filtrations, so the degree is unchanged.

We prove Bernstein inequality for Weyl algebra modules using an injectivity estimate. The faithful finite-step action of a Weyl algebra claim is that the mapis injective for every $j$. Induct on $j$. For $j=0$, a scalar annihilating $M_0\ne 0$ is zero. Suppose $a\in F_j{A}_n$ annihilates $M_j$. For any generator $z=x_i$ or ${\partial }_i$, the commutator $[a,z]$ belongs to $F_{j-1}{A}_n$ and annihilates $M_{j-1}$: both $a(zv)$ and $z(av)$ vanish. The induction hypothesis gives $[a,z]=0$ for every generator. In the ordered basis,In characteristic zero these identities force $a$ to be a scalar. Since it annihilates $M_0$, it is zero. This proves injectivity.

ConsequentlyThe left side grows as a positive constant times $j^{2n}$; the right side has growth degree $2d(M)$. Comparing powers provesThe polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with the usual multiplication and differentiation actions, has $d(M)=n$, showing the bound is sharp.

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.