Ideals of the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$

Question

Let $\mathfrak{g}$ be a Lie algebra, we say that a subvector space $\mathfrak{h}\subset \mathfrak{g}$ is a ideal of $\mathfrak{g}$ if for any $v\in \mathfrak{h}$ and $w\in \mathfrak{g}$ we have $[v,w]\in \mathfrak{h}$.

I'm having troubles to prove that the only ideals of $\mathfrak{sl}_2(\mathbb{C})=\{M\in \mathfrak{gl}_2(\mathbb{C}); \text{Trace}(M)=0\}$ are $\{0\}$ and $\mathfrak{sl}_2(\mathbb{C})$ itself.

I picked $A\in \mathfrak{gl}_2(\mathbb{C})$, an ideal $\{0\}\neq J\subset \mathfrak{sl}_2(\mathbb{C})$ and $M\in J\setminus\{0\}$, then I was trying to find some smart choice of $N\in \mathfrak{sl}_2(\mathbb{C})$ such that $[M,N]=A$, where the Lie brackets here is $[M,N]=MN-NM$. But I'm not finding this $N$.

I think maybe that the "subvector space" hypothesis has to be used, maybe I can suppose that $J$ has dimension $1$ or $2$ and try to get an contradiction, but I'm not sure how to do this.

Answer 1

Let$$X=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad Y=\begin{bmatrix}0&0\\1&0\end{bmatrix},\quad\text{and}\quad H=\begin{bmatrix}1&0\\0&-1\end{bmatrix}.$$Then $\{X,Y,H\}$ is a basis of $\mathfrak{sl}_2(\Bbb C)$. Suppose that $\mathfrak J$ is an ideal of $\mathfrak{sl}_2(\Bbb C)$ and that $\mathfrak j\ne\{0\}$. So, $\mathfrak J$ contains some element $A\in\mathfrak{sl}_2(\Bbb C)\setminus\{0\}$ and then $A=\alpha X+\beta Y+\gamma H$, where at least one of the numbers $\alpha$, $\beta$ and $\gamma$ is not $0$. If $\alpha\ne0$, then $\bigl[Y,[Y,A]\bigr]=-2\alpha Y\in\mathfrak J$, and therefore $Y\in\mathfrak J$. If $\alpha=0$ and $\gamma\ne0$, then $[Y,A]=-2\gamma Y\in\mathfrak J$, and therefore $Y\in\mathfrak J$. And if $\alpha=\gamma=0$, then $\beta\ne0$. So, $\beta Y=A\in\mathfrak J$, and therefore, once again, $Y\in\mathfrak J$.

A similar argument shows that $X\in\mathfrak J$. Since $[X,Y]=H$, $H\in\mathfrak J$ too, and so $\mathfrak J=\mathfrak{sl}_2(\Bbb C)$.