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In an old question, it can be read that "the finiteness of $\text{Gal}(\overline{\mathbf R}/\mathbf R)$" is one of the "impressive finiteness results in mathematics".

I commented the question to know what was meant by the notation $\overline{\mathbf R}$, but I got no answer. Hence my question: what is $\overline{\mathbf R}$, in that context?

For me, this doesn't denote an algebraic closure of $\Bbb R$, because then it would be isomorphic to $\Bbb C$, and the finiteness of $\text{Gal}(\Bbb C/\Bbb R)$ is not hard to establish, in my opinion.

Any comment would be appreciated!

Community
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Watson
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My reading of this quote is that

  • $\overline{\bf R}$ refers to $\Bbb{C}$, and that
  • "impressive finiteness result" refers to the (not that obvious) fact that $\Bbb{C}$ is algebraically closed.

Most of us hear about $\Bbb{C}$ being algebraically closed early in our studies, may be in the same course the complex numbers are first introduced? More often than not the first proof we see is one of the highlights of that sophomore/junior complex analysis course. I am not conversant with the history of FTA, so I don't know if that proof was the first discovered? Surely people had suspected this to be the case much earlier (or were totally uninterested). Personally I have a soft spot for the Galois theoretic proof of $[\overline{\Bbb{R}}:\Bbb{R}]=2$ that only needs the following pieces of analysis:

  • All odd degree polynomials with real coefficients have a real zero.
  • All the complec numbers have a complex square root (ok, this part really only needs trigonometry).

Still, whichever way you look at it, the finiteness of $[\overline{\Bbb{R}}:\Bbb{R}]$ is a non-trivial fact.

Jyrki Lahtonen
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    Thank you! For those who are intersted, here is a Galois theoretic proof of the FTA. Moreover, Artin-Schreier's theorem tells us that fields which are finite subextensions of their algebraic closure behave a little bit like $\Bbb R$. – Watson Jul 23 '16 at 13:11