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?
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?