While recalling some basics of Lie theory, I found a funny proof of the main lemma in Lie's theorem on triangularity of representations of solvable Lie algebras. It turns out that this proof has a very clear and intuitive idea (I wonder why they don't explain it to students), which I present below in a nonrigorous way.
My question is: how, exactly, can you see from the idea that its formalization requires characteristic 0 (or not less than the dimension, or whatever)? What goes wrong with the suggested geometric picture in characteristic $p$? Or what should be added to the informal language I use to make this explicit? Of course I also have a formal version of it, involving characteristic polynomials, and it does spit out some constants that have to be nonzero in the end, but I'm asking for a geometric explanation.
So here it goes. Let $\mathfrak{h} \subset \mathfrak{g}$ be an ideal in a Lie algebra, let $V$ be a finite-dimensional representation of $\mathfrak{g}$, and let $\lambda$ be a weight of its restriction to $\mathfrak{h}$. Then the corresponding weight space $V_\lambda$ is $\mathfrak{g}$-invariant.
"Proof". Clearly everything boils down to showing that $\langle \lambda, [g,h] \rangle = 0$ for all $g \in \mathfrak{g}, h \in \mathfrak{h}$. Now let $\langle \lambda, h\rangle$ be an eigenvalue of $h$ of multiplicity at least $m$, for all $h \in \mathfrak{h}$. Let's extend our ring of scalars by a nilpotent $\epsilon^2 = 0$ to make an infinitesimal perturbation.
On the one hand, $h + \epsilon [g,h]$ is still in $\mathfrak{h}$ (or rather in its extension), since it is an ideal. So it has an eigenvalue $\langle \lambda, h \rangle + \epsilon \langle \lambda, [g,h] \rangle$ of multiplicity $\ge m$.
On the other hand, $h + \epsilon [g, h]$ is conjugate to $h$ (i.e. $h + \epsilon [g, h] = (1 + \epsilon g) h (1 + \epsilon g)^{-1}$), thus it must have the same spectrum as $h$ does. Therefore, in addition to having an eigenvalue $\langle \lambda, h \rangle + \epsilon \langle \lambda, [g,h] \rangle$ it also has an eigenvalue $\langle \lambda, h \rangle$, both of multiplicity $\ge m$. Therefore, either $\langle \lambda, [g, h] \rangle = 0$ for all $h$ (in which case there is nothing left to prove), or $h$ must have had the eigenvalue $\lambda$ with multiplicity $\ge m+1$. Moreover, this actually happens for all $h$. Now run the same argument for $m+1$ instead of $m$, and at some point you will exceed the dimension.
The point of view with nilpotent infinitesimals is not very well known and generally avoided. There is a brief mention if this in James E. Humphreys, Linear algebraic groups, section 9.5 ("dual numbers"), pp. 69-70. I was given this information by Alexandre Borovik.