This is a follow up about a case I'd been meaning to ask in a question I asked about a week ago.
Suppose $V$ is a vector space of dimension $2n$, and let $W(V)$ be the associated Weyl algebra, which can be viewed as an associative $k$-algebra with generators $x_1,\dots,x_n,y_1,\dots,y_n$ satisfying the relations $ [x_i,x_j]=0=[y_i,y_j],\qquad [y_i,x_j]=\delta_{ij}. $
Now let $R=k[X_1,\dots,X_n]$. I'll use $x_i$ to be the $k$-linear operator on $R$ given by multiplication on $X_i$, and let $\partial_i=\frac{\partial}{\partial X_i}$.
I know that there is a homomorphism from the tensor algebra $T(V)\to\operatorname{End}_k(R)$ sending $x_i$ to $x_i$ and $y_i$ to $\partial_i$, which respects the relations above, and hence gives a homomorphism $\varphi\colon W(V)\to\operatorname{End}_k(R)$. In this way $R$ is a $W(V)$ module.
As seen earlier, if $\operatorname{char}(k)=p>0$, then $\partial_i^p=0$, but then $\varphi(\partial_i^p)=y_i^p=0$, so $\varphi$ is not injective with trivial kernel. However, if $\operatorname{char}(k)=0$, does this in fact force $\varphi$ to be a monomorphism? I tried showing the kernel is now trivial, but I'm having a hard time getting a grip on arbitrary $t\in W(V)$ when $\varphi(t)=0$. And does such injectivity imply $R$ is simple and faithful as a module over $W(V)$?