# Is it possible to construct an ordered field which is also algebraically closed?

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.

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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

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.

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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 $x^2 \geq 0$.

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