Complex conjugation in the Galois group of a polynomial

Question

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

Answer 1

Because $P(\bar{z})=\overline{P(z)}$ whenever $P$ has real coefficients therefore if $z$ is a root then $\overline{z}$ is also a root. So if $a$ and $b$ are distinct roots then $\bar{a}$ and $\bar{b}$ are also roots. By the assumption, We can't have four such roots so two pairs of them must be equal. This can either be $a=\bar{a}$ and $b=\bar{b}$ or $a=\bar{b}$ and $b=\bar{a}$. The former means these are real roots, the latter means $a$ and $b$ are conjugate pairs.

Answer 2

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.