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?

• How do you know that $\mathfrak{n}_2 \neq N_{\mathfrak{h}}(\mathfrak{n}_2)$? It's true a posteriori, but I don't see it immediately at that stage of your proof (which otherwise looks ok to me). – Torsten Schoeneberg May 6 at 16:55
• @TorstenSchoeneberg Because a proper subalgebra of a nilpotent algebra is never equal to its own normalizer. – ciccio May 6 at 21:47
• @TorstenSchoeneberg I don't know how to conclude that $ad \; x$ is not nilpotent. – ciccio May 6 at 21:59

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.
• Thank you very much. In the last way can i substitute "$char(\mathbb{F})=0$" with "$\mathbb{F}$ perfect"? – ciccio May 7 at 17:40
• I don't know. One problem is that several equivalent definitions of (semi-)simplicity in characteristic $0$ become inequivalent in positive characteristic, so you would have to say what "semisimple" is supposed to mean now. But probably for none of those possible meanings, $\mathfrak{sl}_2(\Bbb F)$ is semisimple if $char(\Bbb F)=2$ to begin with, regardless of whether $\Bbb F$ is perfect. – Torsten Schoeneberg May 7 at 19:31