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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.

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  • $\begingroup$ You have to exclude $GF(2).$ $\endgroup$
    – gammatester
    Commented Sep 21, 2016 at 12:11
  • $\begingroup$ I don't follow your proof that $1>0.$ Also is there one more order axiom that if $a \neq b$ then either $a<b$ or $a>b$ (but not both)? Because in $Z_2$ interpret $>$ as $\neq$ and then your axioms OR1 and OR2 both hold. $\endgroup$
    – coffeemath
    Commented Sep 21, 2016 at 12:12
  • $\begingroup$ @coffeemath I dont put the definition of an order relation. For $a<b$ we define that $a\le b$ and $a\neq b$. $\endgroup$
    – Masacroso
    Commented Sep 21, 2016 at 12:15
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    $\begingroup$ In step "2", you write "if $a > 0$, ...", but I don't see a demonstration that any such $a$ exists. If you are going to use your OR2, you may be better off trying $a > 0$ and $a^{-1} > 0$ implies $1 = a a^{-1} > 0$ and similarly for $a < 0, a^{-1} < 0$, there there is still some work to do to show either such pair exists. (How do we know it is not the case that for all $a \neq 0$, the sign of $a^{-1}$ is the opposite that of $a$?) $\endgroup$
    – Eric Towers
    Commented Sep 21, 2016 at 12:23
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    $\begingroup$ @Masacroso: The question is about the correctness of the proof, not about the correctness of the statement. $\endgroup$
    – gammatester
    Commented Sep 21, 2016 at 12:36

2 Answers 2

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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.

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  • $\begingroup$ Sorry but I dont understand. If we have a set of cardinality two then $a>b$ and $b>a$ cannot be true at the same time, so the set have a maximum if it have a total order. Maybe I must add this to part 1? $\endgroup$
    – Masacroso
    Commented Sep 21, 2016 at 13:24
  • $\begingroup$ I can't tell if you're commenting on my first paragraph or on the proof that starts in the second paragraph. If the first paragraph, your second note requires $a > b$ implies $a \neq b$. Consequently, your $a_2 > a_1$, $a_3 > a_2$, and $a_1 \neq a_3$ requires three distinct elements. If the proof that follows: No. The argument here uses the pigeon-hole principle. In this case I cannot generate more distinct elements than there are elements in the field, so at some point the list I displayed must repeat an element already on the list. $\endgroup$
    – Eric Towers
    Commented Sep 21, 2016 at 17:04
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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 $i^2$)

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