# “Why” is $[\mathbb{C}:\mathbb{R}] < \infty$?

Obviously this question is a little open-ended.

A lot of complex analysis seems to work primarily because we can view $\mathbb{C}$ as a finite-dimensional $\mathbb{R}$-algebra, and apply analytic and geometric ideas which work only (or at least best) in finite-dimensional real space.

When we consider most other fields that we come across in practice (for instance, $\mathbb{F}_q$,$\mathbb{Q}_p$, or $\mathbb{Q}$) generally their algebraic closures are infinite-dimensional extensions.

Is there any intuitive reason why the way we construct $\mathbb{R}$ would suggest that we were producing a field whose algebraic closure was a finite-dimensional extension? Does such a construction generalize to other fields in any way?

• For one thing, any polynomial of odd degree has a real root... – Grigory M Dec 25 '13 at 10:50
• Yeah, but if that's the only possible answer then the question is somewhat hopeless, since it doesn't seem like there's a way to generalize the IVT to other fields. I was hoping there was a more algebraic explanation. (Assuming that the only proof of that is to observe that an odd-degree function goes to $+\infty$ at one end and $-\infty$ at the other and apply the IVT.) – Daniel McLaury Dec 25 '13 at 10:51
• I wonder if there is any proof that $[\mathbb C : \mathbb R] < \infty$ that does not also prove $[\mathbb C : \mathbb R] = 2$ – GEdgar Dec 25 '13 at 14:27
I can't give you an "intuitive" reason why $\Bbb C$ is a finite dimensional $\Bbb R$-algebra, but I can point you toward a generalization of this fact : $\Bbb R$ is a real closed field, and if $F$ is any real closed field, then $F[\sqrt{-1}]$ is algebraically closed.