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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 $W(V)$?

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Dear Vika, Alex's answer below is correct. To verify the injectivity for yourself, write down a differential operator, and then find a polynomial that is not annihilated by it. If you're not sure how to do that, try first setting $n = 1$ and try low degree operators (degree $0$, $1$, and $2$ to begin with). Once you get a feeling for how the differential operators actually act on various polynomials, you will have no trouble proving the injectivity in general. Regards, –  Matt E Apr 19 '12 at 0:25

1 Answer 1

Yes, this is true. If you are looking at characteristic zero then your map $\varphi$ is an embedding and so $R$ is a faithful $W(V)$ module

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