Let $F$ be a field of characteristic $p > 0$. Let $\mathfrak{g}$ be a linear Lie algebra, that is $\mathfrak{g}\subset M_n(F)$ for some natural number $n$. Does there exist a condition involving $n$ and $p$ such that $\mathfrak{g}$ is semisimple if and only if its Killing form is non-degenerate?
[Math] Semisimplicity of Lie algebra in positive characteristic
lie-algebrasmodular-lie-algebrasrt.representation-theory
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The notions do indeed diverge in positive characteristic: there is the enveloping algebra, and then (in the case that $\mathfrak g$ is the Lie algebra of an algebraic group G) there is also the hyperalgebra of G, which is the divided-power version you mention. In characteristic 0 these two algebras coincide, but in positive characteristic they differ very much. In particular, the hyperalgebra is not finitely-generated in positive characteristic; see Takeuchi's paper "Generators and Relations for the Hyperalgebras of Reductive Groups" for the reductive case. There is also a good exploration of the hyperalgebra in Haboush's paper "Central Differential Operators of Split Semisimple Groups Over Fields of Positive Characteristic."
One can obtain the hyperalgebra as follows. I don't know in what generality the following construction holds, so let's say that $\mathfrak g$ is the Lie algebra of an algebraic group G defined over $\mathbb Z$. Then there is a $\mathbb Z$-form of the enveloping algebra of G (the Kostant $\mathbb Z$-form) formed by taking divided powers, and upon base change this algebra becomes the hyperalgebra. Alternatively, one can take an appropriate Hopf-algebra dual of the ring of functions on G (cf Jantzen's book "Representations of Algebraic Groups").
As for their uses, in positive characteristic the hyperalgebra of G captures the representation theory of G in the way that the enveloping algebra does in characteristic 0, i.e. the finite-dimensional representations of G are exactly the same as the finite-dimensional representations of the hyperalgebra. This fails completely for the enveloping algebra: instead, the enveloping algebra sees only the representations of the first Frobenius kernel of G.
There are many such splittings. Any complementary (aka opposite) parabolic subalgebra (i.e. $\mathfrak{q}$ such that $\mathfrak{p} \oplus \mathfrak{q}^\perp = \mathfrak{g}$ and so on) provides a unique splitting $\mathfrak{p} = \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$. The space of complementary parabolic subalgebras is a $\exp \mathfrak{p}^\perp$-torsor or in other words an affine space modelled on $\mathfrak{p}^\perp$.
We can also equate these to a particular adjoint orbit in $\mathfrak{p}$. There is a particular element $\xi^\mathfrak{p}_\mathfrak{q}$ of $\mathfrak{p}$ called the grading element (aka canonical element) for which $ \mathfrak{q}^\perp \oplus \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$ are the $-1,0,1$ eigenspaces for $\mathrm{ad}\ \xi^\mathfrak{p}_\mathfrak{q}$. This uniquely identifies the complementary subalgebra and the splitting.
Check out this paper on arxiv for an exhaustive treatment of this only assuming the basics of Lie algebras.
Also this paper, covers the filtrations and gradings we are seeing here quite succintly in the preliminary chapter.
Edit: an additional point is that the span of $\xi^\mathfrak{p}_\mathfrak{q}$ is exactly the 1-dimensional centre of $\mathfrak{p} \cap \mathfrak{q}$ and its orthocomplement is semisimple.
Best Answer
The short answer is no. In prime characteristic, the Killing form sometimes behaves badly even for simple Lie algebras. If "semisimple" means that the solvable radical is zero, there is no way to obtain the classical equivalences with non-degeneracy of the Killing form and with the direct sum decomposition into simples. Moreover, the simple Lie algebras have only recently been classified when $p=5$ (Premet-Strade), while for $p=2,3$ little is known and for $p>5$ the classification takes an enormous amount of work (by Block-Wilson and others). Conditions on the dimension of a faithful representation or on the prime are not enough to sort out the concept of semisimplicity. Still, a lot is known. For example, Seligman and others explored in the 1960s the class of modular Lie algebras for which the Killing form is non-degenerate.
ADDED: Much more could be said along these lines, but for older results see the book Modular Lie Algebras by G.B. Seligman (Springer, 1967). Modular Lie algebras have been of much less importance overall than linear algebraic groups, whose Lie algebras are "restricted" (have a nice $p$th power operation) but still don't reflect precisely the group structure or representation theory. Moreover, the representations of simple or more generally semisimple Lie algebras in prime characteristic are poorly understood even for those arising from Lie algebras of semisimple algebraic groups. So working with a fixed $p$ and fixed $n$ will usually not be illuminating.