Prove that a Lie algebra $\mathfrak{g}$ of finite dimension is semi-simple if and only if $\operatorname{rad}(\mathfrak{g})={0}.$ $\newcommand{\rad}{\operatorname{rad}}$I have seen this ($\rad(\mathfrak{g})=0$) is taken as if it were a definition in several places. My definition of Lie semi-simple algebra is as follows:
Def: A Lie algebra is called semi-simple if it is different from $\{0\}$ and it has not  own Abelian ideals.
I have tried to do the following if $\rad(\mathfrak{g})=0$ then as $\rad(\mathfrak{g})$ is the maximal soluble ideal, then $\mathfrak{g}$ has no non-trivial soluble ideals and since the soluble ideals contain the Abelian ideals then there are no Abelian ideals.
I need some help to prove the other direction of this proposition, could someone give me some help or prove everything in a simple way?
 A: Let $\mathfrak{g}$ be a semisimple Lie algebra.  We shall show that $\mathfrak{r}:=\text{rad}(\mathfrak{g})$ is the zero ideal. Since $\mathfrak{r}$ is solvable, the descending series
$$\mathfrak{r}=:\mathfrak{r}^0\supseteq \mathfrak{r}^1\supseteq \mathfrak{r}^2\supseteq \ldots\,,$$
where $\mathfrak{r}^n:=\left[\mathfrak{r}^{n-1},\mathfrak{r}^{n-1}\right]$ for $n=1,2,3,\ldots$, terminates with
$$\mathfrak{r}^M=\mathfrak{r}^{M+1}=\mathfrak{r}^{M+2}=\ldots=0$$
for some integer $M\geq 0$.  If $\mathfrak{r}$ is nonzero, then the minimum possible value $m$ of such an integer $M$ is a positive integer.  Therefore, $\mathfrak{r}^{m-1}$ is nonzero (by minimality of $m$).  However, as $\mathfrak{r}^m=0$, we get
$$\left[\mathfrak{r}^{m-1},\mathfrak{r}^{m-1}\right]=\mathfrak{r}^m=0\,,$$
whence $\mathfrak{r}^{m-1}$ is a nonzero abelian Lie algebra.  However, $\mathfrak{r}^{n}$ is an ideal of $\mathfrak{g}$ for every $n=0,1,2,\ldots$, so we conclude that $\mathfrak{r}^{m-1}$ is an abelian ideal of $\mathfrak{g}$, contradicting the semisimplicity assumption on $\mathfrak{g}$.  Ergo, $\mathfrak{r}=\text{rad}(\mathfrak{g})$ must be zero.
Remark.  As far as I see, the finite-dimensionality assumption on $\mathfrak{g}$ is not needed for the assertion that a Lie algebra $\mathfrak{g}$ is semisimple if and only if $\text{rad}(\mathfrak{g})=0$, provided that the definition of semisimplicity is the lack of nontrivial abelian ideals.  There are other definitions of semisimplicity that may require finite-dimensionality or even that the base field must be of characteristic $0$.
However, there are also different notions of semisimple Lie algebras when the dimension is infinite.  For example, there are terms like "locally solvable Lie algebras" which, to some people, these are Lie algebras whose finite-dimensional subalgebras are always solvable.  In this case, some people define the radical to be the maximal locally solvable ideal, which, to not cause confusion, I shall call the "generalized radical."  To these people, they may define a semisimple Lie algebra to be a Lie algebra whose generalized radical is zero.  To not confuse people here, I will use the term "generalized semisimple Lie algebras" in this case.  Then, my argument above does not seem to work with this definition, if you want to show that a Lie algebra $\mathfrak{g}$ is a generalized semisimple Lie algebra if and only if $\mathfrak{g}$ has no nontrivial abelian ideal.  (The converse of this last statement seems to be false in my opinion.)
