Maximal nilpotent subalgebra which is not a Cartan Subalgebra Consider the special linear algebra $\mathfrak{sl}_2(\mathbb{F}$) and let $\mathfrak{n}_2 \subset \mathfrak{sl}_2(\mathbb{F})$ be the subalgebra of strictly upper triangular matrices. Clearly $\mathfrak{n}_2$ is not Cartan, since it is normalized by elements of the type $\begin{bmatrix}a & b\\ 0 & -a \end{bmatrix}$ which don't lie in $\mathfrak{n}_2$. I would like to show that $\mathfrak{n}_2$ is maximal nilpotent. Clearly it is nilpotent (even commutative). It remains to show that it is maximal. If i am not wrong the normalizer of $\mathfrak{n_2}$ in $\mathfrak{sl}_2(\mathbb{F})$ (that i indicate with $N_{\mathfrak{sl}_2(\mathbb{F})}(\mathfrak{n}_2)$) is $\mathfrak{b}_2 \cap \mathfrak{sl}_2(\mathbb{F})$, where $\mathfrak{b}_2$ is the subalgebra of $\mathfrak{gl}_2(\mathbb{F})$ of upper triangular matrices. If there is another nilpotent subalgebra $\mathfrak{n}_2 \subsetneq \mathfrak{h} \subsetneq \mathfrak{sl}_2(\mathbb{F})$, then i can find an element $x \in \mathfrak{b}_2 \cap \mathfrak{h}$ and $x \notin \mathfrak{n}_2$, since $\mathfrak{n}_2 \subsetneq N_{\mathfrak{h}}(\mathfrak{n}_2)=\mathfrak{h} \cap \mathfrak{sl}_2 \cap \mathfrak{b}_2$. This element will have two distinct eigenvalues (if $\mathrm{char}(\mathbb{F}) \neq 2$). Can i say that the adjoint morphism $ad \; x \colon \mathfrak{h} \to \mathfrak{h}$ is not nilpotent? If it is so, i can conclude by Engel's theorem that $\mathfrak{h}$ is not nilpotent, getting a contradiction.
If what i claim is not true, how to show that $\mathfrak{n}_2$ is maximal?
 A: There are several ways to conclude in the last step.
Maybe the most elegant here is to use that $\mathfrak{b}_2$ is two-dimensional, and that over any field, up to isomorphism there are only two such Lie algebras: The abelian one, and a solvable but non-nilpotent one. See e.g. here and here. It is immediate then that for $char(\Bbb F) \neq 2$, you are in the non-nilpotent case, whereas for characteristic $2$, you are in the abelian case (and indeed $\mathfrak{n}_2$ is not maximal nilpotent).
Another way: Quite generally if $x \in M_n(\Bbb F)$ has characteristic polynomial $\prod_{i=1}^n (X-\lambda_i)$ in an algebraic closure of $\Bbb F$, then $ad(x)$, as endomorphism of $\mathfrak{gl}_n(\Bbb F)$, has characteristic polynomial $\prod_{1\le i,j \le n} (x-(\lambda_i-\lambda_j))$ -- informally speaking, the eigenvalues of $ad(x)$ are the pairwise differences of the eigenvalues of $x$. (This fact can be proven purely computationally; a more conceptual approach is in Bourbaki's volume on Lie groups and algebras, ch. VII §2 no.2 exemple 3.) Now in your case (if $char(\Bbb F) \neq 2$), $x$ has two different eigenvalues, so $ad_{\mathfrak{gl}_2}(x)$ has two non-zero eigenvalues, and since $\mathfrak{gl}_2 = \mathfrak{sl}_2 \oplus \mathfrak{z}(\mathfrak{gl}_2)$, so does its restriction to $\mathfrak{sl}_2$, which therefore cannot be nilpotent.
Another way: if we assume $char(\Bbb F) = 0$, in a semisimple Lie algebra $\mathfrak{g}$, an element $x$ is (ad-)nilpotent iff for every faithful $\mathfrak{g}$-module $M$, the endomorphism $x_M$ induced by $x$ operating on $M$ is nilpotent. Here in particular, taking $M=\Bbb F^2$ implies that if $x$ were ad-nilpotent then the matrix $x$ would be nilpotent itself, which it clearly is not.
