Take an algebraic group $G$ defined over a finite field $K$. Suppose its Lie algebra $\mathfrak{g}$ is simple. It should follow that $G$ is almost-simple. (By this I mean not that $G(K)$ is simple -- I've heard there is a tricky though in some sense inessential counterexample -- but rather that $G$ has no normal algebraic subgroups of dimension greater than $0$ and less than $ \text{dim}\, G(K)$.)

Does anybody know of a concise and straightforward proof of this fact? One can't proceed as for $K=\mathbb{R}$, since $\exp$ is not defined (except on nilpotent subalgebras of $\mathfrak{g}$). I feel this should be simpler than studying the simplicity of $G(K)$. Is this so?

IMPORTANT EDIT: I just realized that I meant to ask the converse of this question. If $G$ is almost-simple in the sense of algebraic groups, how do you prove that $\mathfrak{g}$ is simple? I agree that the direction stated above is very easy.

Yet another important edit: there are exceptions to the converse! See the answers below. They all seem to come from non-trivial centers due to small characteristic. What I really need is the following:

Let $G$ be an almost-simple linear algebraic group. Let $V$ be a subvariety of $G$ going through the origin. Assume $0< dim(V)< dim(G)$. Let $\mathfrak{v}$ be the tangent space to $V$ at the origin. Prove that there is no ideal $\mathfrak{w}$ of $\mathfrak{g}$ containing $\mathfrak{v}$. (Or, what is the same: prove that the conjugates of $g \mathfrak{v} g^{-1}$ of $\mathfrak{v}$ span $\mathfrak{g}$.)

Yet another edit: Aha - not even that is true.

Well, can we at least give a quick proof, without case-work, of the following? Given a classical almost-simple group $G$ (defined as a subgroup of $\SL_n$ by means of equations with integer coefficients), show that there is a constant $c$ such that, for every field $K$ of characteristic $>c$, the Lie algebra $\mathfrak{g}$ of $G(K)$ is simple.

Surely there should be a reduction to the case over $\mathbb{C}$, or a proof from scratch?