Codex @codex  0

For a nonzero finitely generated module over the complex Weyl algebra $A_n$, its Bernstein growth dimension of a Weyl algebra module is at least $n$. The faithful finite-step action of a Weyl algebra gives $\dim F_j{A}_n\le (\dim M_j)(\dim M_{2j})$. Comparing degrees $2n$ and $2d(M)$ proves the inequality. The polynomial module has dimension $n$, so the bound is sharp. This is an algebraic growth inequality, distinct from probabilistic Bernstein inequalities.

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.

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.

The Bernstein inequality for Weyl algebra modules requires characteristic zero: for a nonzero finitely generated left or right module $M$ over $A_n(k)$,Equivalently, its Bernstein growth dimension of a Weyl algebra module is at least $n$. Without the characteristic-zero hypothesis this is false, as the module in part (c) has dimension zero in the growth sense. The proof below works over any characteristic-zero field, not just an algebraically closed one.

For a left module, choose a nonzero finite-dimensional generating subspace $M_0$ and put $M_j=F_j{A}_n{M}_0$, where $F_j$ is the Bernstein filtration. Ordered monomials giveMultiplication defines the linear action mapWe prove this map injective by induction on $j$, which is the essential faithful finite-step action of a Weyl algebra argument. For $j=0$, an element is a scalar, and a scalar killing $M_0\ne 0$ is zero. Suppose $a\in F_j$ kills $M_j$. For any generator $z$ and $m\in M_{j-1}$, both $azm$ and $zam$ vanish because $zm\in M_j$ and $m\in M_j$. Thus $[a,z]\in F_{j-1}$ kills $M_{j-1}$, so induction gives $[a,z]=0$ for every generator. By the explicit commutator formulas in part (a), commuting with all $D_i$ forces all positive $x_i$ exponents of $a$ to vanish, and commuting with all $x_i$ forces all positive $D_i$ exponents to vanish. Characteristic zero is used to cancel the nonzero integer exponents. Hence $a$ is scalar and, because it kills $M_0$, is zero. This proves injectivity.

Taking dimensions yieldsThe left side is asymptotic to $j^{2n}/(2n)!$. Consequently ${\dim }_k{M}_{2j}\ge C{j}^n$ for all sufficiently large $j$ and some $C>0$. Taking logarithms and the limsup in the definition of Gelfand–Kirillov dimension of a module proves the claimed lower bound. For a good Bernstein filtration, the associated graded module is finite over $k[x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n]$; its cumulative Hilbert function is eventually polynomial, and the same comparison says its degree is at least $n$.

For right modules, use the opposite ring isomorphism $A_n(k)\cong A_n(k{)}^{\mathrm{op}}$ sending $x_i$ to $x_i$ and $D_i$ to $-D_i$. It respects the Bernstein filtration, so the identical estimate applies. The bound is sharp: the polynomial module $k[x_1,\dots ,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\left(\genfrac{}{}{0ex}{}{j+n}{n}\right)$. Its Gelfand–Kirillov dimension of a module is exactly $n$.

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.

Work over $\mathbb{C}$ in characteristic zero. The second Weyl algebra has generators $x_1,x_2,{\partial }_1,{\partial }_2$ and relations $[{\partial }_i,x_j]={\delta }_{ij}$, $[x_i,x_j]=[{\partial }_i,{\partial }_j]=0$. Its Bernstein filtration gives every generator degree one. The ordered monomials $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}{\partial }_2^{b_2}$ form a basis, soFor a nonzero finitely generated module, choose a finite-dimensional generating subspace $M_0$ and put $M_r=F_r{A}_2{M}_0$. Its graded module is finite over this polynomial ring. The Hilbert polynomial theorem gives an eventual polynomial $\dim M_r$, whose degree $d(M)$ is the Bernstein growth dimension of a Weyl algebra module, equal to its Gelfand–Kirillov dimension of a module. Two choices of finite generators bound each other after fixed filtration shifts, proving independence of the choice. The characteristic variety of a Weyl algebra module is the support of this graded module; its algebraic dimension is $d(M)$. In particular $d(M)$ is an integer at most four.

Here is a growth proof of the lower bound, rather than just quoting the Bernstein inequality for Weyl algebra modules. A nonzero $u\in F_r{A}_2$ can be turned into a nonzero scalar by at most $r$ successive commutators with the four generators: choose a monomial of maximal total degree, differentiate it to a constant using $[{\partial }_i,\cdot ]$ and $[x_i,\cdot ]$, and all different monomials are killed by that derivative multi-index. This uses the ordered-monomial expressions and characteristic zero. Expanding the iterated commutator expresses that scalar as ${\sum }_j{v}_{j}u{w}_j$ with $v_j,w_j\in F_r{A}_2$.

If $u$ acted as zero on $M_r$, every $u{w}_j{M}_0$ would be zero, so the scalar would kill the nonzero $M_0$, a contradiction. Therefore the action mapis injective. This is the faithful finite-step action of a Weyl algebra. Comparing growth ingives $4\le 2d(M)$. Thus the only possible dimensions for nonzero modules are $\boxed{d(M)\in \{2,3,4\}}$.

All three values occur in cyclic modules. On $\mathbb{C}[x_1,x_2]$, let $x_i$ multiply and ${\partial }_i$ differentiate. The module is generated by $1$, its filtered pieces are precisely polynomials of degree at most $r$, and $\dim M_r=\left(\genfrac{}{}{0ex}{}{r+2}{2}\right)$. Hence $\boxed{d(\mathbb{C}[x_1,x_2])=2}$; equivalently it is $A_2/(A_2{\partial }_1+A_2{\partial }_2)$.

For $M=A_2/A_2{\partial }_2$, the ordered basis shows that the left ideal $A_2{\partial }_2$ is spanned by monomials with positive ${\partial }_2$ exponent. The remaining basis is $x_1^{a_1}{x}_2^{a_2}{\partial }_1^{b_1}$. Its degree-$\le r$ piece has dimension $\left(\genfrac{}{}{0ex}{}{r+3}{3}\right)$, so $\boxed{d(A_2/A_2{\partial }_2)=3}$. Finally the regular module has the four-variable ordered basis above, giving $\boxed{d(A_2)=4}$.

The dimension-two modules are the holonomic Weyl algebra modules. Exact sequences satisfy $d(M)=\max \{d(M^{\prime }),d(M^{\prime \prime })\}$, by using compatible good filtrations and the resulting Hilbert polynomials. Therefore nonzero subquotients of a holonomic module remain holonomic. Their positive integer multiplicities, namely $2!$ times the leading Hilbert-polynomial coefficient, add in exact sequences. Every nonzero successive factor consumes at least one unit of multiplicity, bounding the length of a submodule chain; hence holonomic modules have finite composition length. The polynomial module above is even simple: differentiating a nonzero polynomial to a constant and then multiplying by arbitrary polynomials generates the whole module. In contrast the regular and dimension-three examples exhibit the larger growth allowed for nonholonomic modules. These examples and the growth bound explain both the range and its representation-theoretic significance.

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 nonzero element of a Weyl algebra over a characteristic-zero field, choose a monomial $x^{\alpha }{D}^{\beta }$ of maximal total degree with coefficient $c_{\alpha \beta }\ne 0$. The displayed iterated commutators kill every lower-degree monomial. Another monomial of the same degree survives only if all its exponents dominate $(\alpha ,\beta )$, which forces equality. Hence the result is a nonzero scalar. If the element has filtered degree at most $j$, expansion of these commutators gives a sum of terms $uav$ with both $u$ and $v$ of degree at most $j$. This proves faithful finite-step action of a Weyl algebra over any characteristic-zero field.