# Problem with the definition of simple Lie algebra

I have the following definitions:

Simple Lie algebra: A Lie algebra $\mathfrak{g}$ is simple if it ha no non-trivial (i.e. $0$ or $\mathfrak{g}$ itself) ideals.

Semisimple Lie algebra: A Lie algebra $\mathfrak{g}$ is semisimple if it has no non-trivial (i.e. $0$) solvable ideals.

And the following result:

Claim: A Lie algebra $\mathfrak{g}$ is semisimple if, and only if it is (isomorphic to) the direct sum of simple Lie algebras.

Now I have a problem with this. Indeed let $\mathfrak{g} = \mathbb{C}\oplus \mathfrak{h}$ where $\mathfrak{h}$ is any simple Lie algebra. Then $\mathbb{C}$ is abelian (and thus solvable), and an ideal of $\mathfrak{g}$. At the same time it satisfies the definition of a simple Lie algebra, and so by the claim $\mathfrak{g}$ should be semisimple. Obviously there's something wrong. I think the problem is in the definition of "simple Lie algebra". Can anyone confirm that, or explain where I'm wrong/what I'm not seeing?

• The definition I have before me says a Lie algebra $\mathfrak{g}$ is semisimple iff $\operatorname{rad}\mathfrak{g} = 0$. Commented Aug 24, 2013 at 14:27
• @DanielFischer Yeah, I forgot the "solvable" in "solvable ideals" in the definition. As you can see your definition of semisimple and mine are equivalent, since $\mathrm{rad}\,\mathfrak{g}$ is the maximal solvable ideal of $\mathfrak{g}$. Commented Aug 24, 2013 at 14:33
• One usually requires non-abelian as part of the definition of simple to avoid examples like this. Commented Aug 24, 2013 at 14:42

A simple Lie algebra is a non-abelian Lie algebra whose only ideals are $0$ and itself, see http://en.wikipedia.org/wiki/Simple_Lie_algebra.
The Lie algebras $\mathfrak{g}=[\mathfrak{h},\mathfrak{h}]\oplus \mathfrak{h}=\mathbb{C}\oplus \mathfrak{h}$ are reductive with $1$-dimensional center, e.g., $\mathfrak{gl}_n(\mathbb{C})=\mathbb{C}\oplus \mathfrak{sl}_n(\mathbb{}C)$.