# Why is the Levi-decomposition not available in positive characteristic?

I am learning about Lie-Algebras and I came across a proof of the Levi-decomposition. Can somebody tell why the Lie-Algebra has to be over a field of characteristic $$0$$? Why does it fail for positive characteristic?

Thank you!

The short answer is, that there are counterexamples in characteristic $$p$$, see for example this MO-post. Moreover, most results of classical Lie algebra theory fail in characteristic $$p>0$$, such as Lie's Theorem, Cartan's criterion for semisimplicity (non-degenerate Killing form), almost all results for representations of semisimple Lie algebras, the classification of semisimple Lie algebras and so on.

It is more difficult to give a sufficient answer, why. Of course, many arguments involving linear algebra fail for characteristic $$p$$, e.g., arguments using the trace. A summary can be found, for example, in this post:

If the field of a vector space weren't characteristic zero, then what would change in the theory?

Whitehead's lemma, saying that the $$H^1(L,M)=0$$ for semisimple Lie algebras of characteristic zero also fails in characteristic $$p$$.

In characteristic zero, the short exact sequence

$${\displaystyle 0\rightarrow \mathrm {rad} ({\mathfrak {g}})\rightarrow {\mathfrak {g}}\,{\xrightarrow {\varphi }}\,{\mathfrak {g}}/\mathrm {rad} ({\mathfrak {g}})\rightarrow 0}$$

splits, because $$\mathfrak{g}/{\rm rad}(\mathfrak{g})$$ is semisimple and Whitehead's Lemma. Levi's Theorem is a simple consequence of this fact. As said, this fact fails in characteristic $$p>0$$.