TITLE: Complex conjugation in the Galois group of a polynomial
QUESTION [5 upvotes]: Suppose $P$ is an irreducible polynomial in $\mathbb Q[X]$, with exactly two non-real roots. Then we know these roots must be complex conjugates. 
Why must complex conjugation be an element of $\mathrm{Gal}(P)$?
Thanks

REPLY [4 votes]: The restriction of complex conjugation to a Galois  extension $K$ of $\mathbb Q$ is an automorphism, maybe trivial, of $K$.   [Galois=normal here, because $char.(\mathbb Q)=0$]
This applies to the splitting field $K$ of any polynomial $P\in \mathbb Q[X]$, and solves Matt's problem.      
However this is false if we don't assume that $K$ is Galois.
For example if $\alpha = exp(\frac{2i\pi}{3})\cdot 2^{\frac{1}{3}}$ and $K=\mathbb Q (\alpha)$, the complex conjugate $\bar \alpha$ does not belong to $K=\mathbb Q (\alpha)$, despite a widespread misconception.
Notice that $\alpha$ is a root of $X^3-2\in \mathbb Q[X])$, which proves that Matt's question is not so simple as it looks.