Faithful finite-step action of a Weyl algebra ($F_j{A}_n\hookrightarrow {\mathrm{Hom}}_{\mathbb{C}}(M_j,M_{2j})$)

For $M_j=F_j{A}_n{M}_0$ and $M_0\ne 0$, the action map in the display is injective. If $a\in F_j$ kills $M_j$, each commutator $[a,z]$ with a generator lies in $F_{j-1}$ and kills $M_{j-1}$. Induction makes all these commutators zero, so $a$ is scalar; since it kills $M_0$, it is zero. This finite-step faithfulness does not require the whole module to be finite-dimensional.

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.