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?

  • 2
    $\begingroup$ There are only two possible orderings. Try them. $\endgroup$ – MJD Oct 15 '14 at 15:21
  • $\begingroup$ Well, there's only one field of 2 elements....but here you can use brute force. If there is an order, it is either 0<1 or 1<0. Check your properties that an order must have to see if either of those work $\endgroup$ – Alan Oct 15 '14 at 15:21
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    $\begingroup$ $\underbrace{1+1+1+\dots+1}_n>0$ for any $n$ in an ordered field. Then… $\endgroup$ – egreg Oct 15 '14 at 15:21
  • $\begingroup$ @egreg Umm, no, 1+1=0, which is not >0 :) $\endgroup$ – Alan Oct 15 '14 at 15:22
  • $\begingroup$ @Alan, its a proof by contradiction. Suppose the field can be ordered then $1+1>0$. But as you point out $1+1=0$, a contradiction. $\endgroup$ – goblin Oct 15 '14 at 15:41

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 $0$.

  • $\begingroup$ is it common that in an ordered field that a>0 and b>0 then a+b>0? $\endgroup$ – zhshr Oct 15 '14 at 15:47
  • $\begingroup$ @user2841003 That's part of the definition. $\endgroup$ – egreg Oct 15 '14 at 15:49
  • $\begingroup$ so how about I define that 1<0 or 0<1? In both cases I add 1 to both side and get a contradicting result to the assumption. $\endgroup$ – zhshr Oct 15 '14 at 15:51
  • $\begingroup$ @user2841003 Either $1>0$ or $1<0$; but in the second case $-1>0$, so $(-1)(-1)>0$: a contradiction. So $1>0$. $\endgroup$ – egreg Oct 15 '14 at 15:54

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.

  • 1
    $\begingroup$ It should be $1+1=0$ at the end. $\endgroup$ – egreg Oct 15 '14 at 15:56
  • $\begingroup$ @egreg That's right, will fix. $\endgroup$ – coffeemath Oct 15 '14 at 16:00

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