Lie algebra homomorphism of simple Lie algebra

Question

Let $\mathfrak{g}$ be a simple Lie algebra over $\mathbb{C}$, $\mathfrak{h}$ any Lie algebra over $\mathbb{C}$. Consider $\phi:\mathfrak{g}\rightarrow\mathfrak{h}$ a Lie algebra homomorphism. My question; is $\phi$ an injective Lie algebra homomorphism?

Answer 1

The proof I wrote below is wrong; indeed, a simple Lie algebra has no proper ideals. However, one could have $\mathfrak{i}=\mathrm{Ker}(\phi)=\mathrm{g}$, in which case $\mathfrak{i}\ne 0$, and therefore $\phi$ isn't injective. One could take for example the map $\phi(x)=0\,\forall x\in\mathfrak{g}$, in which case $\mathfrak{i}=\mathrm{Ker}(\phi)=\mathfrak{g}$, and thus $\phi$ isn't injective.

I leave the proof below as I believe it still is of interest.

Let $\mathfrak{i}=\mathrm{Ker}(\phi)\subseteq\mathrm{g}$ denote the Kernel of the homormophism $\phi$; it is an ideal of $\mathfrak{g}$. Indeed, let $x\in\mathfrak{g},\,y\in\mathfrak{i}$. Then one has

\begin{equation} \phi([x,y]_{\mathfrak{g}})=[\phi(x),\phi(y)]_{\mathfrak{h}}=[\phi(x),0]_{\mathfrak{h}}=0, \end{equation}

where $[\,.,.]_{\mathfrak{g}}$ and $[\,.,.]_{\mathfrak{h}}$ denote the Lie bracket in $\mathfrak{g}$ and $\mathfrak{h}$ respectively. Therefore, $[x,y]_{\mathfrak{g}}\in \mathfrak{i}$. As this holds for any $x\in\mathfrak{g}$ and $y\in\mathfrak{i}$, we conclude that $[\mathfrak{g},\mathfrak{i}]_{\mathfrak{g}}\subseteq\mathfrak{i}$, i.e. as claimed the Kernel of a Lie algebra homomorphism is an ideal in $\mathfrak{g}$. As $\mathfrak{g}$ is simple, it has in particular no proper ideals. Therefore $\mathfrak{i}=\mathrm{Ker}(\phi)=0$ -this is wrong; if $\mathrm{Ker}(\phi)=\mathfrak{g}$, then it needs not be $0$ -, and we conclude that $\phi$ is injective. If $\phi$ is also surjective, we can therefore also conclude that $\phi$ is a Lie algebra isomorphism.

However, if given an homorphism and one can find an element $x\in\mathfrak{g}$ such that $x\notin\mathrm{Ker}(\phi)$, then by the above the Kernel must be empty, and $\phi$ therefore injective.