$sl_2(\Bbb R)$ is a simple Lie algebra

Question

Let $D_n$ be the set of derivations of the form $P(t) \frac{d}{dt}$ , where $P\in \Bbb R[t]\setminus \{0\}$ and $deg\ P\le n$.

Let $∂_2 := t^2 \frac{d}{dt}$ and $∂_1 := 2t \frac{d}{dt}$ and $∂_0 = − \frac{d}{dt}$ be elements of $D_2$.

Prove that $sl_2(\Bbb R)$ is simple.

What I did:

I found an isomorphism from $D_2$ to $sl_2(\Bbb R)$, given by:

$f(∂_1) = H, f(∂_2) = X, f(∂_0) = Y $ where :

$H=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, X= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, Y= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$

Now I am stuck with proving that $D_2$ is simple.

Thank you for your help.

Answer 1

Any nonzero ideal of $\mathfrak{sl}_2$ contains one of $H$, $X$ and $Y$ As $[H,aH+bX+cY]=2bX-2cY$ and $[H,[H,aH+bX+cY]]=4bX+4cY$ then if $I$ is an ideal of $\mathfrak{sl}_2$ and $aH+bX+cY\in I$ then $aH\in I$, $bX\in I$ and $cY\in I$.

If $H\in I$ then $[H,X]=2X\in I$ and $[H,Y]=-2Y\in I$ so that $I=\mathfrak{sl}_2$.

If $X\in I$ or $Y\in I$ then $H=[X,Y]\in I$.

All this is in textbooks on Lie algebras.