TITLE: Simple lie algebras, (almost-)simple groups of Lie type
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: Your final request, even if augmented to require the locally closed subvariety $V$ to be smooth, is false over every field $k$ of characteristic 2 and 3, with counterexamples of any rank $n \ge 1$ over any $k$ of characteristic 2, with $V \subset G$ of very large dimension (unbounded as $n$ grows).  So avoiding finite fields of bounded size is insufficient in characteristics 2 and 3.  A source of counterexamples is certain "well-known" exceptional (non-central) isogenies through which Frobenius non-trivially factors, and these isogenies only exist in such characteristics. 
I will describe counterexamples with $G = {\rm{Spin}}_{2n+1}$ for any $n \ge 1$, and a variant of the same ideas (but carried out by more conceptual methods based on the structure theory of semisimple groups rather than based on manipulations in linear algebra) gives counterexamples for ${\rm{Sp}}_{2n}$  ($n \ge 1$) and ${\rm{F}}_4$ over any field of characteristic 2 and for G$_2$ over any field of characteristic 3; see section 7.1 of the book "Pseudo-reductive groups" for a unified development of the relevant general principles
(but again, the phenomenon at hand has been widely known for many decades). 
Let $V = k^{2n+1}$ and
$$q = x_0^2 + x_1 x_2 + \dots + x_{2n-1} x_{2n}.$$
The associated symmetric bilinear form $B_q(v,v') = q(v+v') - q(v)-q(v')$ has defect space $V^{\perp} = \{v \in V\,|\,B_q(v,\cdot) = 0\}$
that is a line since ${\rm{char}}(k)=2$ (this line is $k e_0$). On $V/V^{\perp}$, the induced bilinear form $\overline{B}_q$ is symplectic since ${\rm{char}}(k)=2$. Thus, we get a natural composite homomorphism
$$f: G = {\rm{Spin}}_{2n+1} \rightarrow {\rm{SO}}_{2n+1} \rightarrow
{\rm{Sp}}(\overline{B}_q) = {\rm{Sp}}_{2n}.$$
Let $T$ be a split maximal $k$-torus in $G$, and let $\Delta$ be the base  of $\Phi(G, T)$ associated to a choice of positive system of
roots $\Phi^+$. The multiplication map
$\prod_{a \in \Delta} \mathbf{G}_m \rightarrow T$ defined by
$(\lambda_a) \mapsto \prod a^{\vee}(\lambda_a)$ is an isomorphism since $G$ is simply connected. In particular, ${\rm{Lie}}(T)$ is the direct sum of the "coroot lines" ${\rm{Lie}}(a^{\vee}(\mathbf{G}_m)) = {\rm{Lie}}(a^{\vee})(\partial_t|_{t=1})$.
The kernel $\mathfrak{n} := \ker {\rm{Lie}}(f) = {\rm{Lie}}(\ker f)$ is an ${\rm{Ad}}_G$-stable
subspace of $\mathfrak{g}$, and it is nonzero and proper.  Explicitly, it is the direct sum of the root lines for the short roots and the coroot lines 
${\rm{Lie}}(a^{\vee}(\mathbf{G}_m))$ for the coroots associated to the 
short roots in $\Delta$. (In fact, $\mathfrak{n}$ is the unique
minimal non-central ${\rm{Ad}}_G$-stable subspace of $\mathfrak{g}$.) Hence, $\mathfrak{n} = {\rm{T}}_e(V)$ for the smooth locally closed subvariety $V \subset G$ through the identity $e$ given as the closed subvariety of the open cell associated to $\Phi^+$ by forming the direct product (embedding via multiplication into $G$) of the $T$-root groups for the short roots and the direct factors $a^{\vee}(\mathbf{G}_m)$ for short $a \in \Delta$.