Are soluble/nilpotent lie algebras always isomorphic to a subalgebra of upper triangular matrices? For soluble lie algebras, every representation has some basis such that the image is a subalgebra of upper triangular matrices (if you assume the field is algebraically closed). Then by Ado's theorem, over (an algebraically closed field of) characteristic zero, any soluble lie algebra is a subalgebra of some upper triangular matrices.
For nilpotent lie algebras, (it seems to me) you can't quite apply the same idea directly, because for a representation of a nilpotent lie algebra, you can only conclude there is a basis where the image is represented by strictly upper triangular if you additionally assume that all elements act by nilpotent endomorphisms too (but here you don't need to assume the field is algebraically closed). However, there is a form of Ado's theorem for nilpotent lie algebras, which says that over characteristic $0$, you can always find a faithful representation such that this additional assumption is true, and so nilpotent lie algebras over characteristic $0$ are a subalgebra of some strictly upper triangular matrices.
Firstly, I would like to know if this reasoning is correct and if so that this is the easiest way of getting to these conclusions.
Secondly, it seems that Ado's theorem holds for characteristic $p$ by a proof by Iwasawa. Is there a good reference for this? Can you still take the faithful representation of a nilpotent lie algebras to act by nilpotent endomorphisms?
If not then if $k$ has characteristic $p$, is there a nilpotent lie algebra not isomorphic to any subalgebra of strictly upper triangular matrices?
Additionally, if $k$ is not algebraically closed of arbitrary characteristic, then is there a soluble lie algebra not isomorphic to any subalgebra of upper triangular matrices? I'm sure that Lie's Theorem doesn't hold over non-algebraically closed fields, but that doesn't mean that this is automatically the case.
Thanks, please let me know if I'm missing anything obvious!
 A: I don't follow your particular reasoning, but here's a proof (Ado being granted) that every nilpotent Lie algebra in char zero can be represented by strictly upper triangular matrices, over a field $K$ of characteristic zero.
First suppose $K$ is algebraically closed. Using Ado, take a faithful $\mathfrak{g}$-module $V$. Then (see e.g. Bourbaki), $V=\bigoplus V_\chi$, where $\chi$ ranges over $\mathrm{Hom}(\mathfrak{g},K)$ (Lie algebra homomorphisms) and $V_\chi=\{v\in V:\forall g\in\mathfrak{g}:(g-\chi(g))^{\dim(V)}v=0\}$. That is, on $V_\chi$, the action is scalar + nilpotent, the scalar being given by $\chi$. Then we can modify the module structure on $V_\chi$, to remove the scalar (i.e., act with the same matrices, but with zero diagonal). The resulting representation is unchanged on $[\mathfrak{g},\mathfrak{g}]$, hence its kernel has trivial intersection with $[\mathfrak{g},\mathfrak{g}]$. Hence, adding an abelian nilpotent action, we can produce a faithful nilpotent module.
Next, if $K$ is arbitrary (of char. zero), start with a faithful representation; on a finite extension $L$ of $K$ it can be made upper triangular, and the previous argument applies. Eventually we have a faithful nilpotent $L\otimes\mathfrak{g}$-module, and by (Weil) restriction of scalars this yields a faithful $\mathfrak{g}$-module.

I don't think this adapts to char $p$. However, let me emphasize that Ado in characteristic $p$ is known as much easier than its char 0 counterpart (due to the enveloping algebra being PI). I think it's covered in Jacobson's book. I'd actually guess that the proof actually provides directly a nilpotent module when the Lie algebra is nilpotent.
