Ideals of $\mathfrak{gl}_n$

Question

How does one determine the ideals of $\mathfrak{gl}_n(C)$?

My guess is that the only ones are $(0) $ and $\mathfrak{sl}_n(C)$.

I think approaching the problem by the fact that each $\mathfrak{g}^{ (i)}$ is an ideal of $\mathfrak{g}$ would be wise.

Answer 1

Your guess is correct, with the exception that also the center is always a Lie ideal. Two days ago we have determined the center of $\mathfrak{gl}_n(K)$ here, for a field $K$ of characteristic zero. Indeed, the center is $1$-dimensional, spanned by $I_n$. Since $\mathfrak{gl}_n(K)\cong \mathfrak{sl}_n(K)\oplus K$, and $\mathfrak{sl}_n(K)$ is semisimple, the only ideals of $\mathfrak{gl}_n(K)$ are $$ (0), \; K, \, \mathfrak{sl}_n(K),\; \mathfrak{gl}_n(K). $$

Answer 2

$gl_n(C)$ is an example of reductive Lie algebra. You can write $gl_n(C)=sl_n(C)\oplus C$. Since $sl_n(C)$ is semi-simple. The only ideals of $gl_n(C)$ are $sl_n(C)$ and $C$.