Let $F$ be an algebraically closed field and let $L$ be a simple Lie algebra of dimension $n$ over $F$. Let $ad: L\longrightarrow End\_F(L)$ denote the adjoint representation of $L$. If $F$ has characteristic zero, then it is well-known that, for every linear transformation $f\in ad(L)$, the semisimple and nilpotent parts of $f$ are also contained in $ad(L)$ (see e.g. Section 5.4 in the Humphreys's book "Introduction to Lie algebras and representation theory").


In general, the same conclusion is not true if $F$ has characteristic $p>0$. Is it possible to have the extreme case in which $ad(L)$ does not contain any nonzero semisimple linear transformation of the vector space $L$? 



Simple Lie algebras and Jordan decomposition
Any finite dimensional simple Lie algebra over an algebraically closed field of characteristic $p>3$
contains a nonzero element $x$ such that $ad(x)$ is semisimple. This is a nontrivial fact, and the only proof I can think of relies on the classification. For $p>3$, the simple Lie algebras split
into three families: Lie algebras of simple algebraic groups and their quotients, filtered Lie algebras of Cartan type, and the Melikian algebras (which only occur when $p=5$). The first class is easy to sort out, of course. The third class has a very explicit presentation which is easy to find in the literature. Then one can see straight away that each Melikian algebra $\mathcal{M}(m,n)$ admits two commuting linearly independent elements $t\_1,t\_2$ sich that $ad(t\_i)^p=ad(t\_i)$ for $i=1,2$. 


The real issue with this problem (as well as with many similar problems) is the existence of filtered Lie algebras of Cartan type $H$ corresponding to Hamiltonian forms of the first kind. Such Hamiltonian forms involve parameters which are sometimes organised as sets of matrices in canonical Jordan form. Sadly, the commutator relations in the corresponding Lie algebras depend on these parameters as well, which makes computations an unpleasant experience.


However, the good news is that any finite dimensional simple Cartan type Lie algebra $L$ contains a unique maximal subalgebra of smallest codimension. It is called the $standard\ maximal\ subalgebra$ of $L$ and often denoted $L\_{(0)}$. Skryabin proved in [Comm. Algebra, 23 (1995), 1403-1453] that under our assumptions on $p$ the subalgebra $ad(L\_{(0)})$ of $\mathfrak{gl}(L)$ is closed under taking $p$-th powers; see Theorem 2.1 in $loc.\ cit$. 
By maximality, $L\_{(0)}$ does not consist of $ad$-nilpotent elements. Since $ad(L\_{(0)})$ is closed under taking $p$-th powers, it is straightforward to see that there is a nonzero element $x\in L\_{(0)}$ such that $ad(x)$ is semisimple (in fact, one can say a lot more than that).


I have no idea as to what happens when $p$ is $2$ or $3$ as we have no classification in these characteristics. This adds to a long list of open problems of which my favourite is: does there exist a finite dimensional simple Lie algebra $L$ with a finite automorphism group. This never happens when $p>3$ and there is a rather short proof of this fact in the literature.


