I am trying to understand a definition in chapter $1$ of Humphreys Lie Algebra. I think I need the following result to understand the definition properly:
Let $\mathcal L$ be a non zero Lie Algebra over some (algebraically closed) field $F$. Recall that $a \in \mathcal L$ is called ad-nilpotent if $ad(a): \mathcal L \to \mathcal L$ is a nilpotent endomorphism. Is it true that every non zero Lie algebra has an ad-nilpotent element?
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For more general fields see Varea, "EXISTENCE OF AD-NILPOTENT ELEMENTS AND SIMPLE LIE ALGEBRAS WITH SUBALGEBRAS OF CODIMENSION ONE", Proc. A.M.S. 104 (1988), 363-368.
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Here's an argument (with no use of the solvable radical and semisimple quotient).
First, if $\mathfrak{g}$ (finite-dimensional) admits a grading in $\mathbf{Z}$, then every element of $\mathfrak{g}_n$ for $n\neq 0$ is ad-nilpotent (clear).
So a Lie algebra with the property that no nonzero element is ad-nilpotent has no grading in $\mathbf{Z}$. In general this can happen (e.g., the real Lie algebra $\mathfrak{so}(n)$ for $n\ge 3$). However, if the field is algebraically closed of characteristic zero, this implies that the Lie algebra is nilpotent. But then every element is ad-nilpotent; since it was assumed that no nonzero element is ad-nilpotent, this implies that the Lie algebra is zero.
Now let us check that $\mathfrak{g}$ is nilpotent (assuming the field algebraically closed). Indeed, any 1-dimensional multiplicative group of automorphisms yields such a nontrivial grading. Hence if there's no nontrivial grading in $\mathbf{Z}$ (by trivial grading I mean the grading concentrated in degree 0), the automorphism group of $\mathfrak{g}$ is virtually unipotent. In particular, the Lie algebra of derivations is nilpotent. Hence the Lie algebra of inner derivations is nilpotent, which in turn implies that $\mathfrak{g}$ is nilpotent.
I'm curious about this argument in positive characteristic (although the conclusion is known with an elementary proof Benkaart and Isaacs, 1977). Namely, does there exist a non-nilpotent finite-dimensional Lie algebra, in positive characteristic (over an algebraically closed field), with no nontrivial grading in $\mathbf{Z}$ (or equivalently, with $\mathrm{Aut}(\mathfrak{g})_0$ unipotent)? Added: This is precisely Question (c) in Section 7 of D. Winter, On groups of automorphisms of Lie algebras, J. Algebra 8, 131-142 (1968).
Here is a sketch of the proof in the finite-dimensional case. You start with the Levi-Malcev decomposition of your Lie algebra ${\mathfrak g}= {\mathfrak r} \oplus {\mathfrak s}$, where ${\mathfrak r}$ is the solvable radical and ${\mathfrak s}$ is the semisimple part. Note that the action of ${\mathfrak s}$ on ${\mathfrak r}$ is, in general, nontrivial. Next, you use the Gauss decomposition ${\mathfrak s}= {\mathfrak n}_+ \oplus {\mathfrak t} \oplus {\mathfrak n}_-$, where ${\mathfrak n}_\pm$ are nilpotent subalgebras of ${\mathfrak s}$. If ${\mathfrak s}\ne 0$, so are ${\mathfrak n}_\pm$. Furthermore the adjoint action of ${\mathfrak n}_\pm$ on ${\mathfrak r}$ is also nilpotent. (This is a general fact about semisimple Lie algebras: Nilpotent elements have nilpotent action under any representation.) Thus, if ${\mathfrak s}\ne 0$, then ${\mathfrak g}$ contains nonzero nilpotent elements (any nonzero element of ${\mathfrak n}_\pm$ would do). Consider now the case when ${\mathfrak s}=0$, ${\mathfrak g}={\mathfrak r}$. Then you use the further decomposition ${\mathfrak r}= {\mathfrak n}\oplus {\mathfrak t}$, where ${\mathfrak t}$ is abelian and ${\mathfrak n}$ is the nilponent radical of ${\mathfrak r}$. The subalgebra ${\mathfrak n}$ consists of nilponent elements. Moreover, ${\mathfrak r}$ admits a faithful linear representation where ${\mathfrak t}$ maps to the diagonal matrices and ${\mathfrak n}$ maps to strictly upper triangular ones. In any case, each nonzero element of ${\mathfrak n}$ gives a nonzero nilpotent element of ${\mathfrak r}$. Lastly, if ${\mathfrak n}=0$ then ${\mathfrak g}$ is abelian and every element is nilpotent (by the definition).