Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is well known that while the real numbers are totally ordered, they are not algebraically closed, and while the complex numbers are algebraically closed, they are not totally ordered. Is it possible to construct a totally ordered algebraically closed field? If so, an example would be appreciated.

Thanks in advance!

share|improve this question
    
Please don't include signatures in your posts. See the FAQ. –  Arturo Magidin Nov 25 '11 at 5:42
    
Sorry about that, won't happen again. –  Dan M. Katz Nov 25 '11 at 5:45

2 Answers 2

up vote 14 down vote accepted

No, we cannot do it.

No field of positive characteristic can be totally ordered: since $1$ is a square, it must be greater than $0$, and then so are $1$, $1+1$, $1+1+1$, etc; but if the characteristic is $p\gt 0$, then $1+1+\cdots+1$ with $p$ summands is both positive and equal to $0$ which is impossible.

So a totally ordered field must be of characteristic zero. In particular, $-1\neq 1$.

Since $1\gt 0$, then $-1\lt 0$. If your field contains a root of $x^2+1$, call it (for lack of a better name) $i$, then $i^2=-1$; but this means that $-1$ is positive, because it is a square. This contradicts the fact that $-1$ is negative. So no totally ordered field can contain a root of $x^2+1$, let alone be algebraically closed.

share|improve this answer
    
Cool stuff...Thanks! –  Dan M. Katz Nov 25 '11 at 5:59
    
@user6300, more cool stuff: if an ordered field $F$ "loks like" the reals then $F(\sqrt{-1})$ is algebraically closed, just like $\mathbb R$ and $\mathbb C$. Look up real closed field, which are discussed nicely in Jacobson and Lang and others. –  lhf Nov 25 '11 at 12:17

Suppose by contradiction that you have such an order.

Pick an element $a<0$. The equation $x^2=a$ has a root. But then $x<0$, $x=0$ or $x>0$, all implying that $x^2 \geq 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.