Show that a finite field cannot be ordered Check my proof please.
*First note that ordered here mean totally ordered.
**Second note. For $a>b$ I mean that $a\ge b$ and $a\neq b$.
1) Any finite set totally ordered have a maximum. Proof: suppose that a total ordered set dont have maximum, i.e. for any $a\in K$ exists some $b\in K$ such that $a<b$.
Then for some $a_1$ exists $a_2>a_1$, for $a_2$ exists $a_3>a_2$ with $a_3\neq a_1$, because if $a_3=a_1$ then we will have that $a_1>a_2>a_1\implies a_1>a_1$ due to the transitive property of any order relation.
In general if $a_k>a_{k-1}$ then $a_k\notin\{a_1,...,a_{k-1}\}$.
If $|K|=n$ then we have that for $a_n$ must exist some $a_k$ such that $a_k>a_n$, but this contradicts the rule that $a_k\notin \{a_1,...,a_n\}=K$. Then exists a maximum for a finite set totally ordered.
2) If $K$ is an ordered field then $1>0$. The order axioms of an ordered field are:


*

*OR1: if $a>b$ then $a+c>b+c$ for any $a,b,c\in K$.

*OR2: if $a>0$ and $b>0$ then $ab>0$ for any $a,b\in K$.
Theorem TH1: if $\{0,1\}$ is a field then it cannot be ordered. Proof: if $1>0$ then


*

*$1+1=1$ then $1+1>1+0\iff 1>1$ what is a contradiction.

*$1+1=0$ then $1+1>1+0\iff 0>1$ what contradicts OR1.
If we suppose that $0>1$ instead $1>0$ we get similar statements that deny the possibility that the field $\{0,1\}$ can be ordered.
Theorem TH2: exist some $a>0$ with $a\notin\{0,1\}$ in any ordered field. Proof: we have that $aa^{-1}=1$ for any $a\neq 0$ from the axioms of field. 
Then $aa^{-1}+(-1)=0$.


*

*Suppose that $1>0$ then $aa^{-1}>0$ and then from OR2 we get that $a>0$.

*Suppose that $1<0$ then $aa^{-1}<0$ but then $(-a)(-a)^{-1}>0$, where we get that $(-a)>0$.

*In any case exists some $a>0$ in any ordered field provided that from TH2 we see that any ordered field must have cardinality greater than $2$.
Then if $a>0$ and because $a=a\cdot 1$ due to OR2 we have that $1>0$.
3) Any finite field cannot be ordered. If $K$ is finite and totally ordered then $K$ have a maximum, i.e. exists some $M\in K$ such that $M\ge a$ for all $a\in K$.
Then $M+1\le M$, but we have that 


*

*if $M+1=M$ then $(-M)+M+1=(-M)+M$, but then we get that $1=0$ what cannot be true if $K$ is a field ($1\neq 0$ id an axiom for the definition of a field).

*if $M+1<M$ then $(-M)+M+1<(-M)+M$ what implies that $1<0$, what cannot be possible. 
So we conclude that a finite field cannot be ordered.
 A: Another viewpoint for the same topic, just for general information.
If a field is ordered then all sum of squares $\sum a_i^2$ equal to $0$ implies each $a_i=0$. From this you easily get that $-1$ can be a sum of squares in all finite field precisely because it is not ordered. For example in $\Bbb F_7$ you have $-1=2^2+3^2$.
In the important case of $\Bbb C$ it is not ordered because $-1$ is a square (equal to $i^2$)
A: After your second note, I have to agree with @gammatester's comment that your part 1 uses three unique elements.  I don't see how to fix this for three general elements, but we need not work with general elements...
Move your part 2 to the front, giving $0<1$.  Then construct the characteristic subring  $$  0 < 1 < 1+1 < 1+1+1 < 1+1+1+1 < \cdots,  $$
which sequence must eventually repeat an element because the field is finite.  (Precisely, $(\mathrm{char}(K) - 1)+1$ is the first repeated element for fields of positive characteristic.)  Then you have an element strictly less than itself in the order.
