Let L be a Lie algebra with a onedimensional maximal subalgebra. Is the following true? Over a perfect field of characteristic 0 or p > 3, every such finitedimensional Lie algebra is either 2dimensional, or is a form of sl(2). General structure theory seems to indicate this.

$\begingroup$ As the answer by Yves suggests, it's risky here to work over a field which isn't algebraically closed. Probably that will cause problems also in prime characteristic, but I'm not sure. $\endgroup$ – Jim Humphreys Jul 29 '13 at 20:05

$\begingroup$ @Jim: I think the main issue is certainly the characteristic: issues such that I mention are not a big deal. Also the obvious 3dim examples disappear when the field is algebraically closed, at least in char 0. $\endgroup$ – YCor Jul 29 '13 at 20:53

$\begingroup$ The only possibilities over an algebraically closed field are two dimensional, so the question then is no longer interesting. $\endgroup$ – David Towers Jul 30 '13 at 10:34
If $\mathfrak{a}$ is 1dimensional and $\mathfrak{v}$ is an irreducible $\mathfrak{a}$module then $\mathfrak{a}$ is maximal in $\mathfrak{a}\ltimes\mathfrak{v}$. Such $\mathfrak{v}$ of dimension $\ge 3$ indeed exists if the ground field has extensions of degree $\ge 3$, and then this Lie algebra has dimension $\ge 4$ with a maximal 1dimensional subalgebra. (I guess this are the only finitedimensional counterexamples, at least in characteristic zero).

$\begingroup$ Of course. Thank you Yves. That probably does sort out the characteristic zero case. $\endgroup$ – David Towers Jul 29 '13 at 17:15

1$\begingroup$ Yes: assume $\dim(\mathfrak{g})\ge 3$, $\mathfrak{a}$ is 1dim and maximal. If $\mathfrak{g}$ is not simple and $\mathfrak{s}$ is a minimal ideal, then $\mathfrak{g}=\mathfrak{a}\ltimes\mathfrak{s}$ by maximality. If $\mathfrak{s}$ is abelian it has to be irreducible as in my example. Otherwise it's simple and since derivations are inner the semidirect we see that $\mathfrak{a}$ is not maximal, contradiction. Otherwise, $\mathfrak{g}$ is simple. [...] $\endgroup$ – YCor Jul 29 '13 at 20:58

1$\begingroup$ [...] if $\mathfrak{g}$ is simple, its rank (in the sense of Bourbaki Lie Algebras Chap 7) is 1. This passes to the "complexification" (which is semisimple) and thus $\mathfrak{g}$ is 3dimensional by the classification, since the Bourbaki rank is the usual (absolute) rank of semisimple Lie algebras. [If $\mathfrak{g}$ is a $n$dimensional Lie algebra over an infinite field, its Bourbaki rank is the minimal dimension of $Ker(ad(x)^n)$ when $x$ ranges over the Lie algebra.] $\endgroup$ – YCor Jul 29 '13 at 21:04

$\begingroup$ Yes, if the characteristic of F is zero and g is simple then g is threedimensional and \sqrt{F} \not \subseteq F. When char F > 7 then the results of Benkart and Osborn apply, but I'm not sure about characteristics 5 and 7. $\endgroup$ – David Towers Jul 30 '13 at 9:15

$\begingroup$ @David: but, well, I'm not sure what's going on with this argument in positive characteristic: if you have a simple Lie algebra $\mathfrak{g}$ and $K$ is the algebraic closure of the ground field, what can be said of $\mathfrak{g}\otimes K$? is it a product of simple Lie algebras? I can't even prove that having a trivial solvable radical passes to nonseparable extensions, but I'm not a specialist. $\endgroup$ – YCor Jul 30 '13 at 22:06