# Let $M$ be a subfield of Complex field such that $M/\Bbb Q$ is a finite Galois extension

Let $$M$$ be a subfield of Complex field such that $$M/\Bbb Q$$ is a finite Galois extension. Show that if $$[M:\Bbb Q]$$ is an odd number, then $$M$$ is a subfield of Real field.

My current thought is since $$M/\Bbb Q$$ is Galois, it is a normal extension. That is, if $$f(x)$$ has a root in $$M$$, all roots of $$f(x)$$ are in $$M$$. However, I don't know how to continue from here.

## 1 Answer

Since $$M/\mathbb{Q}$$ is finite and Galois, it can be thought of as the splitting field of some separable polynomial $$p(x)$$ with rational coefficients. If $$M$$ were not a subfield of $$\mathbb{R}$$, then $$p(x)$$ would have a nonreal complex root, $$\alpha$$. Since all of $$p$$'s coefficients are real then $$\overline{\alpha}$$ would be a root of $$p$$ as well. Now complex conjugation would be a nontrivial automorphism of $$M$$ fixing $$\mathbb{Q}$$, so it would be in the Galois group. But complex conjugation has order 2 as an automorphism, so the size of the Galois group, $$[M:\mathbb{Q}]$$ would be even by Lagrange's theorem.