= Solution
The <Bernstein inequality for Weyl algebra modules> requires characteristic zero: for a nonzero finitely generated left or right <module> $M$ over $A_n(k)$,
$$
\boxed{\operatorname{GKdim}_{A_n(k)}M\ge n.}
$$
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 $M_j=F_jA_n\,M_0$, where $F_j$ is the <Bernstein filtration>. Ordered <monomials> give
$$
\dim_kF_jA_n=\binom{j+2n}{2n}.
$$
Multiplication defines the linear action map
$$
F_jA_n\longrightarrow\operatorname{Hom}_k(M_j,M_{2j}),\qquad a\longmapsto(m\mapsto am).
$$
We 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 $M_0\ne0$ 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 yields
$$
\binom{j+2n}{2n}\le(\dim_kM_j)(\dim_kM_{2j})\le(\dim_kM_{2j})^2.
$$
The left side is asymptotic to $j^{2n}/(2n)!$. Consequently $\dim_kM_{2j}\ge Cj^n$ 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 $k[x_1,\ldots,x_n,\xi_1,\ldots,\xi_n]$; 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> $k[x_1,\ldots,x_n]$, generated by $1$, has $M_j$ equal to the polynomials of degree at most $j$, of dimension $\binom{j+n}{n}$. Its <Gelfand–Kirillov dimension of a module> is exactly $n$.
Back to article page
