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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.

Srivatsan
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IBS
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  • Can someone find all the sub Lie algebra of W_1, or simply Der(F[x]) (derivation Lie algebra of polynomials algebra in one varialbe)? I think all the sub Lie algebra of Der(F[x]) is the following three types: (1) dim=1: spanned by $u\partial_x$ for any $u\in F[x])$; (2) dim=2: spanned by $x\partial_x, x^2 \partial_x$; (3) dim=3 spanned by $\partial_x, x\partial_x, x^2\partial_x$. That is to say dim>4 is imposible. Can you prove it? – hnzt Dec 24 '11 at 11:12
  • I don't know if this is all of them but, for any $k$, $x^k \partial_x$, $x^{k+1} \partial_x$, ..., $x^{p-1} \partial_x$, span a subalgebra. In particular, once $p$ is $>5$, you can get dimensions greater than $4$ in this way. – David E Speyer Dec 24 '11 at 12:03
  • The span of $x^1 \partial_x$, $x^3 \partial_x$, $x^5 \partial_x$ etcetera is also a subalgebra. At this point, I don't believe there is going to be a simple description. This really deserves to be a separate question, and it would probably also benefit from you thinking about it some more. – David E Speyer Dec 24 '11 at 12:25
  • It seems not easy, the two element $x^3\partial_x,x^5\partial_x$ when p=7 can also span a subalgebra. – hnzt Dec 24 '11 at 15:28
  • Please turn these two answers of yours into a question. Hopefully you will gather sufficiently many points so as to be able to post comments. – Mariano Suárez-Álvarez Dec 24 '11 at 18:47

1 Answers1

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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.

David E Speyer
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  • Thanks. I managed to do it all for $n=1$. My problem is actually the generalization – IBS Dec 03 '11 at 15:30
  • Well, have you figured out the analogue of Step 1 yet? – David E Speyer Dec 03 '11 at 15:51
  • Well I understand that if $\partial$ is our derivation, it's enough to look at $\partial\left(x_{i}\right)$ for every $i$, But this gives us $n$ different $u$ as in Step 1... – IBS Dec 03 '11 at 16:03