Simplicity of $\operatorname{Der} \left(\mathbb F_p [x_1, \dots, x_n]/ (x_1^p,\dots, x_n^p )\right)$ I need to prove that the Lie algebra defined as: $W_{n} = \operatorname{Der} \left(\mathbb F_p [x_1, \dots, x_n ] / (x_1^p, \dots, x_n^p )\right)$, when $(x_1^p, \dots, x_n^p )$ is the ideal generated from the monomials $x_1^p, \dots, x_n^p$ and when $p$ is prime, is simple. 
I have tried following several strategies, such as taking a nontrivial ideal and trying to prove it will be all, but couldn't do it. 
Also, I know that this Lie algebra is bounded, and I tried using this fact, but without success. 
I would greatly appreciate any help!
Thanks.
 A: Since this is labeled as homework, here are hints. I'll give some guide posts for the case $n=1$.
Step 1: Every derivation of $\mathbb{F}_p[x]/x^p$ is of the form $f \mapsto u f'$, for some $u \in \mathbb{F}_p[x]/x^p$. Here $f'$ is the standard derivative.
Call this derivation $u \partial_x$.
Step 2: What is the commutator $[\partial_x, u \partial_x]$?
Step 3: Let $I$ be a nonzero ideal, so $I \ni u \partial_x$ for some nonzero $u$. Then $I$ contains $[\partial_x, u \partial_x]$, $[[\partial_x, [\partial_x, u \partial_x]]$, etcetera. Show from this that $\partial_x \in I$.
Step 4 Now, with $I$ as above, show that $I$ is the whole Lie algebras of derivations.

Okay, now for more than one variable. 
Step 1' Every derivation is of the form $g \mapsto \sum u_i \partial f/\partial x_i$, for some $n$-tuple of polynomials $u_i$. 
We will call this derivation $\sum u_i \partial_i$.
Step 2' What is $[ \partial_j, \sum u_i \partial_i]$?
Step 3' So what sort of element can we conclude that every ideal contains?
Step 4' Finish the proof.
