How to prove a finite field is not ordered?

Question

I have a set S={0,1}, and the addition and multiplication rules are \begin{array}{c|cc} +&0&1\\ \hline 0&0&1\\ 1&1&0 \end{array}

\begin{array}{c|cc} *&0&1\\ \hline 0&0&0\\ 1&0&1 \end{array}

It is sure that it is a field. How can I prove this field can be ordered or not?

Answer 1

An ordered field $F$ must have characteristic $0$, because $$ \underbrace{1+1+\dots+1}_n > 0 $$ for all $n>0$. A finite field can't have characteristic $0$.

Answer 2

If it is ordered, that means there is a nonempty subset $P$ such that (1) for each $x$ exactly one of $x \in P, x=0, -x\in P$ (2) If $x,y \in P$ then each of $x+y,x*y$ is in $P$. There are not many choices for $P$ here. $P$ cannot have $0$ in it, otherwise (1) above is false. And $P$ is nonempty. So far the only possible $P$ is $P=\{1\}.$ But this $P$ does not satisfy (2) since $1 \in P$ but $1 + 1=0$ is not in $P.$ [Thanks to @egreg for noticing my previous $1*1=0$ version wasn't right.]

Added: An easier way to see it can't be ordered: Assume that $a<b$ is incompatible with $b<a$, and that one can always add the same thing to both sides of an inequality. Then since $1 \neq 0$ we have either $0<1$ or else $1<0$ (but not both). However adding $1$ to each side of either of these gives the other, i.e. a contradiction.