Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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?

share|cite|improve this answer
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
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.