= Solution
Work over $\mathbb C$, or another <field> of <characteristic zero>. The <Weyl algebra> $A_n$ is generated by $x_1,\ldots,x_n,\partial_1,\ldots,\partial_n$ 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> is
$$
F_mA_n=\operatorname{span}\{x^\alpha\partial^\beta:|\alpha|+|\beta|\le m\},\qquad\operatorname{gr}A_n\cong\mathbb C[x_1,\ldots,x_n,\xi_1,\ldots,\xi_n].
$$
Thus $\dim F_mA_n=\binom{m+2n}{2n}$. <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 $M_m=\sum_jF_mA_nm_j$ 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. Define
$$
\operatorname{Ch}(M)=\operatorname{Supp}(\operatorname{gr}M)=V(\operatorname{Ann}\operatorname{gr}M)\subseteq\mathbb A^{2n}\cong T^*\mathbb A^n.
$$
This 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 $M_i\subseteq\widetilde M_{i+c}$ and conversely, since each finite set of generators occurs in bounded degree in the other filtration. If a degree-$d$ operator has symbol annihilating $\operatorname{gr}M$, 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 $\operatorname{gr}\widetilde M$. 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 <polynomial>
$$
\dim M_m=\frac{e(M)}{d!}m^d+O(m^{d-1}),\qquad d=\dim\operatorname{Ch}(M).
$$
Here $d$ 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 $d\le2n$.
Here is a proof sketch of the essential lower bound, rather than an appeal to its name. For a nonzero $f\in F_mA_n$, choose a nonzero <monomial> of greatest total degree in its ordered expansion, say $x^\alpha\partial^\beta$. The operations $\operatorname{ad}\partial_i$ and $\operatorname{ad}(-x_i)$ 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 gives
$$
1=\sum_j u_jfv_j,\qquad u_j,v_j\in F_mA_n.
$$
Choose $0\ne v\in M_0$. If $f$ acted as zero on all $M_m$, every $fv_jv$ would vanish, contrary to $v=\sum_ju_jfv_jv$. Hence the linear action map
$$
F_mA_n\longrightarrow\operatorname{Hom}_{\mathbb C}(M_m,M_{2m})
$$
is injective. It follows that $\dim F_mA_n\le(\dim M_m)(\dim M_{2m})$. Comparing <polynomial> degrees gives $2n\le2d$. Thus
$$
\boxed{n\le d(M)\le2n}\qquad(M\ne0).
$$
This proves <Bernstein inequality for Weyl algebra modules> through the <faithful finite-step action of a Weyl algebra>.
The bound is sharp. The <polynomial> representation $\mathbb C[x_1,\ldots,x_n]\cong A_n/\sum_iA_n\partial_i$ 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> $A_n/\sum_iA_nx_i$ 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 $n\ge1$, 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.
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