Abstract Algebra – 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.