Ideals of $\mathfrak{sl}_n(k)$ .

PROPOSITION 5.22   Let $k$ be a field of characteristic $p$ (possibly 0 .) Let $n$ be a positive integer.

1. If $p\not \vert n$ , then $\mathfrak{sl}_n(k)$ is a simple Lie algebra.
2. If $p \vert n$ , then $\mathfrak{sl}_n(k)$ has a unique nontrivial ideal $k. 1_n$ .

PROOF.. Let $I$ be an ideal of $\mathfrak{sl}_n(k)$ . By taking trace we see immediately that

$$1_n\in \mathfrak{sl}_n(k) \iff p\vert n.$$

Thus if $I\subset k.1_n$ , then The only nontrivial possibility is that $p \vert n$ and $I=k.1_n$ .

Assume now that $I\not \subset k.1_n$ . Then by an argument similar to that in Proposition 5.19, we see that

$$x=c_0 1_n+c_1 e_{1 n} \in I, \quad (\exists c_0,c_1\in k, c_1\neq 0)$$

holds.

(1) If $p\not \vert n$ , then by taking trace we see that $e_{1 n} \in I$ . By permuting the basis, we see that $e_{i j}\in I$ whenever $i\neq j$ . Thus $I=\mathfrak{sl}_n(k)$ in this case.

(2) If $p \vert n$ , then by assumption on $(p,n)$ we have $n\geq 3$ . Thus

$$I \ni [e_{2 1}, x]=c_1 e_{2 n} \quad \therefore e_{1 n}=[e_{1 2} ,e_{2 n}]\in I.$$

So in this case also we see that $I=\mathfrak{sl}_n(k)$ .

$\qedsymbol$

PROPOSITION 5.23   Let $k$ be a field of characteristic $2$ . Then any two dimensional Lie algebra $L_{[b:c]}$ as in Lemma 5.20 is an ideal of $\mathfrak{sl}_2(k)$ . Thus each ideal $I$ of $\mathfrak{sl}_2(k)$ is equal to the one in the following list.

1. $0.$
2. $k.1_n.$
3. $L_{[b:c]}$
4. $\mathfrak{sl}_n(k)$

PROOF.. Easy exercise. $\qedsymbol$