Ideals of the Lie algebra $\mathfrak{sl}_2(\mathbb{C})$ [duplicate]
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 $\mathfrak J=\mathfrak{sl}_2(\Bbb C)$.