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Let $L= \mathfrak{sl}(3, F)$, with $charF= 3$. I want to prove that $L/Z(L)$ is semisimple. I already know that $Z(L)$ is the set of scalar matrices in $L$. I think it is sufficient to show that every solvable ideal, or equivalently, the radical lies in $Z(L)$, but I don’t know how to move on from here. Any hints?

cip
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  • It is easy to show that $L/Z(L)$ has no proper nonzero ideal (short direct computation). So it is a simple Lie algebra, hence semisimple. – Dietrich Burde Oct 27 '20 at 19:46
  • So by picking $x \in I$, with with $I$ being a nonzero ideal of $L/Z(L)$ and $x$ being a non scalar matrix, the goal is to show that $[x,y] \in L/Z(L), \forall y \in L/Z(L)$. I guess my problem now is to find a basis for $L/Z(L)$ so that I can express $x$ and $y$ in terms of that basis and use bracket rules: how is it the basis of $L/Z(L)$ related to the basis of $L$? – cip Oct 28 '20 at 09:02
  • Just mod out the $1$-dimensional center, so if your basis was ${e_1,\ldots ,e_8}$ and $Z=\langle e_8\rangle$, then set $e_8$ to zero. But recall that $3=0$. – Dietrich Burde Oct 28 '20 at 09:49
  • Starting from the standard basis of $L$ made of $h_1, h_2, e_{ij}$ with $i \neq j$, the basis of $L/Z(L)$ is made of $h_1, e_{ij}$. If I show that $e_{ij} \in I$, I’m done because $h_1 = [e_{12}, e_{21}]$. Now $x \in I$, so it is a linear combination of $L/Z(L)$ basis, so I can suppose $x = e_{ki}$ and compute $[e_{ki}, e_{ij}] = e_{kj} \in I$. Is this correct? – cip Oct 28 '20 at 12:04
  • By the way, the Lie algebra $L/Z(L)$ is called $\mathfrak{psl}(3)$. It is $7$-dimensional. If $I$ is a nonzero ideal, you have to show that $I=\mathfrak{psl}(3)$. I don't see that this follows if $h_1\in I$? Recall that $h_1=\operatorname{diag}(1,-1,0,0,0,0,0)$. – Dietrich Burde Oct 28 '20 at 12:06
  • I want to show that all elements of the basis of $L/Z(L)$ belong to $I$. I do it first for elementary matrices, then because $h_1$ is a bracket of elementary matrices, I get that $h_1$ also belongs to $I$ – cip Oct 28 '20 at 12:22
  • I basically applied the same reasoning normally used to show that $\mathfrak{sl}(2)$ is simple – cip Oct 28 '20 at 12:43
  • $h_2$ is an element of the basis of $L$, but not of $L/Z(L)$, which is 7-dimensional. In fact I noticed that $Z(L)$ is generated by $h_1 - h_2$ which is linearly indipendent from $h_1$. That's why I concluded that the basis of $L/Z(L)$ is $h_1, e_{ij}$ with $i \neq j$ – cip Oct 28 '20 at 15:45
  • Sorry to bother again @Dietrich Burde: do you think my argument is correct? – cip Oct 29 '20 at 09:09
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    Yes, this is the right way to do it. There is an alternative way (see the duplicate). Show that there are no $2$-dimensional subalgebras. Then the algebra is automatically simple. – Dietrich Burde Oct 29 '20 at 09:10

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