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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

$\displaystyle 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

$\displaystyle 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

% latex2html id marker 7437
$\displaystyle 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$


next up previous
Next: Invariant bilinear forms and Up: generalities in finite dimensional Previous: Ideals of
2007-12-19