adjoint representation is irreducible iff $\mathfrak{g}$ is simple

I am trying to prove that for a Lie algebra $$\mathfrak{g}$$:

$$ad_{\mathfrak{g}}$$ the adjoint representation of $$\mathfrak{g}$$ is irreducible iff $$\mathfrak{g}$$ is simple.

I tried to use the fact that stable ideals of $$ad_{\mathfrak{g}}$$ are ideals of $$\mathfrak{g}$$: Then if $$\mathfrak{h}$$ is a stable space under $$\mathfrak{g}$$ it is $$\{0\}$$ or the entire $$\mathfrak{g}$$. but I Couldn't go further.

Subrepresentations of the adjoint representation just correspond to ideals of $$\mathfrak{g}$$. Since $$\mathfrak{g}$$ is simple, they are trivial or the whole Lie algebra. Hence the adjoint representation is irreducible.
• Thank you! This is my attempt to prove the correspondence: $\mathfrak{h}$ ideal of $\mathfrak{g} \Leftrightarrow [\mathfrak{h},\mathfrak{g}] \subset \mathfrak{h} \Leftrightarrow \forall X\in \mathfrak{g}\: ad_X(\mathfrak{h}) \subset \mathfrak{h} \Leftrightarrow \mathfrak{h}$ is subrepresentation of $ad_{\mathfrak{g}}$. is that correct? it's quite obvious but just to be sure I get it. – PerelMan Apr 15 at 17:35