I know that this question has already answers but I would like to get better and therefore don't want to look at a solution. I don't want full solutions. My question was if my choice of a maximal element is valid. I'm new to this subject and I think that it's valid to choose an maximal element. But I don't trust my handwaving and want to be sure. My proof attempt:
Assume that there is a finite ordered field K. Then since K is ordered and finite there exists a $m \in K$ ,$a<m$ for every $a\in K\(a)$ Since K has at least two elements, 0 and 1, the neutral elements of addition and multiplication, we have $0<1 \leq m$. We know that $x,y \in K$, $x,y>0$ fulfill $x+y>0$ since $K$ is ordered. Take $x=y=m$. Then $2m>0$ . Because m is maximal $m>2m$. But this is equivalent to $m<0$. That contradicts our assumption and therefore there is no such field.
