About the Center of the Special Linear Group $SL(n,F)$

I want to classify the center of the Special Linear Group. I already determined the center for $SL(n,F)$:

$$Z(SL(n,F))=\left\{ \lambda I_n:\lambda^n=1 \right\}.$$

I showed that $Z(SL(n,F))$ is itself a group and now

I want to show that $Z(SL(n,F))$ is cyclic and has a order dividing $n$.

I thought this is possible by regarding the map $SL(n,F)\rightarrow P(F)$, $P(F)$ being the projective space. The kernel of this map is $Z(SL(n,F))$. How do I have to argue now?

You have already proved that $Z\bigl(SL(n,F)\bigr)$ is (isomorphic to) the group $\{\lambda\in F\mid\lambda^n=1\}$. And this group is cyclic, since finite subgroups of the multiplicative group of a field are cyclic. Extending $F$ to its algebraic closure $\overline F$, you'll get a group with $n$ elements (the $n$ roots of the polynomial $x^n-1$). Since your group is a subgroup of this one and this one has order $n$, the order of your group divides $n$, by Lagrange's theorem.