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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 $\varphi\colon W(V)\to\operatorname{End}_k(R)$.

I'm curious about the injectivity of $\varphi$ depending on the characteristic of the field $k$. If $\operatorname{char}(k)=p>0$, by Alex Youcis' answer, I see that $\partial^p_i=0$, but does this somehow imply $\varphi$ is not injective?

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up vote 1 down vote accepted

It's pretty obvious why $\partial_i^p=0$. Just take a general element and apply $\partial_i^p$ $p$ times. Any terms with less than $p$ powers of $x_i$ are killed as constants and anything with $p$ or more is killed because you will be multiplying by $p$.

Now, suppose that the homo $T(V)\to\text{End}_k(R)$ was injective, then $T(y_i^p)=T(y_i)^p=\partial_i^p=0$ would imply that $y_i^p=0$--is it?

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Dear Alex, do you mind clarifying your last point? I'm not quite following what you say, since this seems to apply to the map $T(V)\to\operatorname{End}_k(R)$, and not $\varphi\colon W(V)\to\operatorname{End}_k(R)$. –  Vika Apr 14 '12 at 21:27
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@Vika: Dear Vika, There are some typos in Alex's answer, which are easily fixed. Firstly, he wrote $T(V) \to \mathrm{End}_k(R)$ when he meant $W(V) \to \mathrm{End}_k(R)$. Secondly, he is then using $T$ to denote this homomorphism. In short, he has shown that $y_i^p$ maps to zero. Of course, $y_i^p$ is not zero in $W(V)$. QED. Regards, –  Matt E Apr 17 '12 at 2:08
    
Dear @MattE, thank you for taking the time to clarify this. –  Vika Apr 17 '12 at 6:35
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