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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.

For a filtered algebra $A$ with finite-dimensional filtration pieces, a good filtration of a module on a finitely generated module is compatible with multiplication and has finitely generated associated graded module over the associated graded ring. Taking bounded-degree translates of a finite set of module generators supplies such a filtration of a module when the associated graded ring is Noetherian.

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.

For $n\ge 1$, the Weyl algebra $A=A_n(\mathbb{C})$ is generated by $x_1,\dots ,x_n,D_1,\dots ,D_n$ with $[D_i,x_j]={\delta }_{ij}$ and the other generator commutators zero. Iterating the differential Ore extension construction gives the ordered basis $x^{\alpha }{D}^{\beta }$. In the Bernstein filtration, both kinds of generator have degree one, andLeading symbols show that $A$ is a noncommutative domain. The Hilbert basis theorem and homogeneous-generator lifting by decreasing degree make it both left and right Noetherian.

It is also a simple ring. In a nonzero two-sided ideal, choose a nonzero element and a monomial $x^{{\alpha }_0}{D}^{{\beta }_0}$ of maximal total degree with nonzero coefficient $c$. Applying $(\mathrm{ad}D{)}^{{\alpha }_0}(\mathrm{ad}x{)}^{{\beta }_0}$ eliminates every lower-degree term and every different term of that maximal degree. The result is the nonzero scalar $(-1{)}^{|{\beta }_0|}{\alpha }_0!{\beta }_0!c$. Thus the ideal contains $1$. Here the scalar extraction by iterated Weyl commutators uses characteristic zero. Despite simplicity, $A$ has no nonzero finite-dimensional modules: in such a representation $[D_1,x_1]=I$ would have trace $0=\dim M$, impossible over $\mathbb{C}$. Its interesting module theory is necessarily infinite-dimensional.

For a finitely generated left module $M$, choose a finite-dimensional generating subspace $M_0$ and set $M_j=F_{j}A{M}_0$. This is a good filtration of a module. Its associated graded module is finite over the graded polynomial ring, so its cumulative Hilbert function is eventually a polynomial,with positive integer multiplicity $e(M)$. These are the Bernstein growth dimension of a Weyl algebra module and its multiplicity of a filtered module. Two good filtrations bound each other after fixed shifts: lift finite homogeneous generators in one filtration, bound their degrees in the other, and use compatibility with $F_{j}A$; then reverse the roles. The growth degree and leading coefficient therefore do not depend on the good filtration of a module. The degree is also the Gelfand–Kirillov dimension of a module and the dimension of the characteristic variety of a Weyl algebra module, the support of $\mathrm{gr}M$. Since $M$ is a quotient of a finite sum of copies of $A$, $d(M)\le 2n$.

The essential lower bound is the Bernstein inequality for Weyl algebra modules:Here is a full proof. Multiplication gives a linear map $F_{j}A\to {\mathrm{Hom}}_{\mathbb{C}}(M_j,M_{2j})$. Prove it injective by induction on $j$. At $j=0$ a scalar killing $M_0\ne 0$ must vanish. If $a\in F_{j}A$ kills $M_j$, then for every generator $z$, its commutator $[a,z]\in F_{j-1}A$ kills $M_{j-1}$: both $azm$ and $zam$ are zero for $m\in M_{j-1}$. Induction gives $[a,z]=0$ for all generators. In the ordered basis, $[D_i,a]$ differentiates the $x_i$ exponents and $[x_i,a]$ differentiates the $D_i$ exponents with a minus sign. Vanishing of all these commutators in characteristic zero forces $a$ to be a scalar. It then vanishes because it kills $M_0$. This proves the faithful finite-step action of a Weyl algebra and henceComparison of degrees gives $2n\le 2d(M)$, proving the inequality. Equally, without invoking eventual polynomiality, this estimate directly bounds the Gelfand–Kirillov dimension of a module below by $n$. Right modules satisfy the same inequality through the filtration-preserving isomorphism $A\cong A^{\mathrm{op}}$, $x_i\mapsto x_i$, $D_i\mapsto -D_i$.

A nonzero finitely generated module with $d(M)=n$ is a holonomic Weyl algebra module; the zero module is conventionally included. This minimal-growth class is stable under submodules, quotient modules and extensions. For $0\to N\to M\to Q\to 0$, take the induced and quotient filtrations. Their graded sequence is exact; Noetherianity of $\mathrm{gr}A$ makes the induced filtrations good. Thus the Hilbert functions add andif both nonzero terms have the top dimension, their multiplicities add. For a holonomic Weyl algebra module, every nonzero subquotient has dimension $n$, by the upper bound inherited from $M$ and Bernstein's lower bound. Each nonzero factor in a strict chain therefore consumes positive integral multiplicity. This proves the multiplicity bounds the length of a holonomic Weyl module:In particular a holonomic module is both Noetherian and Artinian and has finite length, although its vector-space dimension is infinite. This distinction separates finite-dimensional representation theory from finite-length Weyl module theory.

The polynomial module $\mathbb{C}[x_1,\dots ,x_n]$, with $x_i$ acting by multiplication and $D_i$ by differentiation, is generated by $1$. Its $M_j$ consists of polynomials of degree at most $j$, with dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$, so $d=n$ and $e=1$. It is simple: differentiating a maximal-degree monomial of a nonzero polynomial isolates a nonzero constant, and multiplication by polynomials then generates the whole module. A second basic example is the delta moduleUsing the reverse ordered basis $D^{\beta }{x}^{\alpha }$, its vector space is $\mathbb{C}[D_1,\dots ,D_n]$; $D_i$ acts by multiplication and $x_i$ by negative differentiation. The same argument proves simplicity and holonomicity. These two examples are exchanged by the automorphism $x_i\mapsto D_i$, $D_i\mapsto -x_i$. Free modules, in contrast, have growth dimension $2n$ and are not holonomic. Thus the central organizing invariants are good filtration of a module, growth dimension, characteristic support and multiplicity, with Bernstein's estimate explaining why dimension $n$ is the natural smallest nonzero size.

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.

Choose a finite-dimensional generating vector subspace $V\subseteq A$ containing $1$, and write $V^n$ for the linear span of products of $n$ elements of $V$. The Gelfand–Kirillov dimension isFor a nonzero finitely generated module $M$ considered as a right module, choose a finite-dimensional generating vector subspace $W\subseteq M$ and define the Gelfand–Kirillov dimension of a module byFor left modules, replace $W{V}^n$ by $V^{n}W$. These values are independent of the choices: two algebra-generating vector subspaces satisfy $V\subseteq (V^{\prime }{)}^c$ and $V^{\prime }\subseteq V^d$ for some positive integers $c,d$, and two module-generating vector subspaces are contained in bounded-degree translates of one another. The corresponding growth bounds differ only by a constant rescaling and shift of $n$, which do not change the limit superior.

Let $S$ be a nonzero standard graded algebra, generated by $r$ elements of degree one. It is a homogeneous quotient ring of the polynomial ring $k[z_1,\dots ,z_r]$. By the Hilbert-Serre theorem, its Hilbert series is a rational function whose only possible pole is at $t=1$. After canceling, write it as $P(t)/(1-t{)}^d$ with $P(1)\ne 0$, where $0\le d\le r$. The cumulative dimensions are the coefficients of $P(t)/(1-t{)}^{d+1}$ and are eventually a polynomial in $n$ of degree $d$, with positive leading coefficient. With $V=S_0\oplus S_1$, these cumulative dimensions equal $\dim V^n$. ConsequentlyFor $d=0$ the algebra is finite-dimensional and nonzero, so the cumulative dimension is eventually a positive constant. If the zero algebra is allowed, the common convention $\mathrm{GKdim}0=-\infty$ is an exception to the printed assertion; the integer conclusion concerns nonzero unital algebras.

For the Weyl algebra $A=A_1(k)=k⟨X,D⟩/(DX-XD-1)$, the total-degree filtered algebra structure has associated graded ring $\mathrm{gr}A=k[x,\xi ]$. Its ordered monomials $X^i{D}^j$ form a basis. Every finitely generated module has a good filtration of a module, and its associated graded module is finitely generated over $k[x,\xi ]$. The Hilbert-Serre theorem therefore shows that its Gelfand–Kirillov dimension is an integer in $\{0,1,2\}$.

A nonzero module of Gelfand–Kirillov dimension zero here would have eventually constant cumulative dimension, hence be finite-dimensional. This is impossible in characteristic zero: taking the trace of the endomorphisms representing $DX-XD=1$ gives $0={\dim }_{k}M$. For right modules the right-action operators reverse composition and give the negative identity instead, with the same contradiction. ThusBoth occur. The regular module $A$ has $\dim V^n=\left(\genfrac{}{}{0ex}{}{n+2}{2}\right)$, hence Gelfand–Kirillov dimension two. For a right module of Gelfand–Kirillov dimension one, take $M=k[x]$ with $f\cdot X=xf$ and $f\cdot D=-f^{\prime }$. These actions satisfy the defining relation because $(f\cdot D)\cdot X-(f\cdot X)\cdot D=f$. The module is a cyclic module generated by $1$, and $\dim (1\cdot V^n)=n+1$. This also gives the lower bound for this particular Weyl algebra without invoking a general inequality for higher Weyl algebras.