Center of $\mathfrak{sl}(n,F)$

Prove that $\mathfrak{sl}(n,F)$ (matrices with trace zero) has center $0$, unless $\operatorname{char}F$ divides $n$, in which case the center is $\mathfrak{s}(n,F)$ (scalar multiples of the identity).

I don't understand why the center can be $0$. Aren't scalar multiples of the identity always in the center, because they commute when multiplied by any other matrix? ($AB=BA$ if $A$ is a scalar multiple of the identity.)

Solution 1:

We can show the claim either by direct computation or by Lie algebra theory. In the first case let $E_{ij}$, $i\neq j$ and $H_i=E_{ii}-E_{i+1,i+1}$ for $i=1,\ldots n-1$ be a basis of $L={\frak{sl}} (n,k)$ and $A\in Z(L)$. Then $0=[A,E_{ij}]=AE_{ij}-E_{ij}A$ for all $i\neq j$ and $0=[A,H_i]=AH_i-H_iA$ for all $i$ imply that $A=0$, if $p\nmid n$.

On the other hand, we can use that $L$ is simple in characteristic zero. This means, that every ideal, in particular the center $Z(L)$, is zero or ${\frak{sl}}(n,F)$ itself. However $L=Z(L)$ would imply that $L$ is commutative, which means $n=1$ and $Z(L)=\frak{sl}(1)=0$. In the other case, $Z(L)=0$.
In characteristic $p$ this argument is more complicated, because for $p\mid n$ the center is in fact $1$-dimensional and $L$ then is not simple. However, $L/Z={\frak{psl}}(n,F)$ then is simple for $n>2$.