"Almost-ideals" in the (simple) Lie algebra of an algebraic group?

Question

Let $G<\mathrm{GL_n}$ be a simple linear algebraic group defined over a finite field $K$. Let $\mathfrak{g}$ be its Lie algebra. Assume $\mathfrak{g}$ is simple.

Is it necessarily the case that there is no subspace $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}_g$ for every $g\in G(K)$?

Note: it is clear that there is no $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}$ for every $g\in G(\overline{K})$. It is also clear that the answer to the question above is "yes" when the number of elements of $K$ is larger than a constant depending only on $n$: since $\mathfrak{v}$ is not an ideal, there is a $v\in\mathfrak{v}$ such that all $g\in G(\overline{K})$ such that $\mathrm{Ad}_g(v)\in \mathfrak{v}$ lie in a proper subvariety of $G$.

Note 2: A friend has just proposed over the breakfast table that there are linear algebraic groups with no non-trivial rational points over $K$. That would obviously imply an answer of "no" to my question. I am not convinced that such a thing is really possible, at least not when we are talking about the group $G(K)$, $G$ simple (as opposed to more exotic groups of Lie type). EDIT: as per Venkataramana's comment below, this situation cannot, in fact, occur for $G$ simple.

Answer 1

You are looking for the following theorem of Steinberg:

Let q=|K|. An irreducible algebraic representation of $G(\overline{K})$ of highest weight λ remains irreducible when restricted to G(K) if $\langle \lambda,a^\vee\rangle <q$ for all simple coroots $\alpha^\vee$.

The reference is here. When G is simply connected, Steinberg proves more, namely that all irreducible modules for G(K) arise from this construction.

In the situation at hand, we need to apply Steinberg's theorem to the adjoint representation. There are only a small number of cases where the condition $\langle \lambda,a^\vee\rangle <q$ does not apply, easily classified on a case by case basis. I haven't checked, but I would suspect that in each of these cases, $\mathfrak{g}$ is not simple.