A colleague of mine asked me the question below, and since I could not answer it, I thought I might have more luck on MO.


In Encyclopedia of Mathematics, a finite dimensional Lie algebra $L$ over a field$\newcommand{\bK}{\mathbb{K}}$ $\bK$ is defined to be *supersolvable* if all eigenvalues of all operators $\DeclareMathOperator{\ad}{ad}$ $\ad(X)$ belong to $\bK$. It is then stated that any supersolvable Lie algebra over a field of characteristic $0$ can be isomorphically imbedded in the Lie algebra of upper-triangular matrices with coefficients from $\bK$. Does anyone know a proof or a reference for the last statement?


**Edit 1.** Here is the precise reference: *[Lie algebra, supersolvable. V.V. Gorbatsevich (originator), Encyclopedia of Mathematics](http://www.encyclopediaofmath.org/index.php?title=Lie_algebra,_supersolvable&oldid=17913).*


**Edit 2.** Let me refine the original question following YCor's suggestions. Let $\frak{g}$ be a Lie algebra over $\mathbb{R}$. Let’s call $\frak{g}$ *supersolvable* if it admits a complete flag made up of ideals. (By the way, Knapp calls such algebras *split-solvable*.)


Gorbatsevich-Onishchik-Vinberg state without a proof (see Lie groups and Lie algebras III, p.50) that if $\frak{g}$ is supersolvable, then it is triangulable. Any idea how to prove it?


Clearly not every faithful representation of a supersolvable $\frak{g}$ is triangulable, even for abelian $\frak{g}=(\mathbb{R},+)$, so Ado’s theorem is not enough.



About supersolvable Lie algebras
I've checked the published book ([(on Springer's site)](http://www.springer.com/us/book/9783540546832) *Lie groups and Lie algebras III* by Gorbatsevich-Onishchik-Vinberg, from which the site linked by Liviu most likely refers.


The book contains a wealth of results, but with few proofs, and indeed makes the mistake spotted by Dave. So let me clarify the picture.


Let $K$ be a field. Let's say that a finite-dimensional Lie algebra $\mathfrak{g}$ over $K$ is


* pointwise ad-triangulable if for every $x\in\mathfrak{g}$, all eigenvalues of $\mathrm{ad}(x)$ are in $K$;
* supersolvable if it admits a complete flag made up of ideals
* triangulable if it is isomorphic to a Lie subalgebra of the algebra of upper triangular matrices in large enough dimension.


Immediate implications:


* supersolvable implies solvable and pointwise ad-triangulable
* triangulable implies supersolvable;


If $K$ has characteristic zero, the first implication is an equivalence:


* (char$(K)=0$) solvable and pointwise ad-triangulable implies supersolvable.


Indeed if $\mathfrak{g}$ is abelian there is nothing to do, otherwise the center $\mathfrak{z}$ of $[\mathfrak{g},\mathfrak{g}]$ (which is nilpotent; I possibly use char. zero here) is nontrivial and the adjoint action of $\mathfrak{g}$ on $\mathfrak{z}$ factors through an abelian group and has an common eigenvector as it consists of a commuting family of trigonalizable operators. [This argument works with no characteristic assumption if $\mathfrak{g}$ is assumed nilpotent-by-abelian from the beginning, i.e. has nilpotent derived subalgebra.]


* So I guess the post should be corrected as whether (in char. zero) supersolvable implies triangulable. Actually in the above-quoted book, they say that it can be proved by following the proof of Ado's theorem (which is quite complicated). [But as I mentioned as a common, it is not enough to just evoke Ado's theorem and then use the Lie-Kolchin theorem, because the representation produced by Ado's theorem could fail to be conjugate to a triangular representation, even when we start from an abelian Lie algebra.]
* Finally, we can wonder about the implication whether pointwise ad-triangulable implies supersolvable, which, in characteristic zero (as I henceforth assume), is, by the above, equivalent to wonder whether pointwise ad-triangulable implies solvable. As I said, this implication is claimed to always hold but Dave mentioned that it clearly fails when the field is algebraically closed. The authors were misled by the fact they had the real case in mind, in which case the implication does hold. To clarify, let me state the general case:



> 
> **Proposition.** Let $K$ be a field of characteristic zero. Equivalences: (i) Every finite-dimensional pointwise ad-triangulable Lie algebra over $K$ is solvable (and hence supersolvable). (ii) $K$ admits a field extension of degree 2.
> 
> 
> 


Examples of fields with no extension of degree 2 (so that pointwise ad-triangulable fails to imply solvable) are not only algebraically closed field, but also, for instance, the field generated by iterated square roots of rationals. On the other hand the reals, number fields, and finite extensions of $p$-adic fields, admit extensions of degree 2.


*Proof of the proposition:* assume that $K$ admits an extension of degree 2 (i.e. admits a non-square $t$). It's enough to show that there is no simple Lie algebra $\mathfrak{g}$ over $K$ that is pointwise ad-triangulable. Indeed, since $\mathfrak{g}$ admits a Cartan subalgebra defined over $K$, we immediately see that $\mathfrak{g}$ is $K$-split so admits $\mathfrak{sl}\_2(K)$ as a subalgebra. Notice that $\mathfrak{sl}\_2(K)$ contains an element $x$ such that ad$(x)$ is not trigonalizable: just pick $x=\begin{pmatrix}0 &t\\ 1 & 0\end{pmatrix}$; then the characteristic polynomial of ad$(x)$ (in $\mathfrak{sl}\_2(K)$) is $X(X^2-4t)$, which is not split since $t$ is not a square, and in $\mathfrak{g}$, the characteristic polynomial of ad$(x)$ is divisible by $X(X^2-4t)$, so is not split. Thus $\mathfrak{g}$ is not pointwise ad-triangulable.


Conversely if every element in $K$ is a square, then for every $x\in\mathfrak{sl}\_2(K)$, the characteristic polynomial $\chi(X)$ of ad$(x)$ has degree 3 and is divisible by $X$, and hence is split (since the degree 2 quotient $\chi(X)/X$ is then split using that the discriminant is a square); thus in this case $\mathfrak{sl}\_2(K)$ is pointwise ad-trigonalizable but not solvable. 


