# Complex conjugation in the Galois group of a polynomial

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

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Isn't complex conjugation an automorphism of the complex numbers as a field? – Andrés Caicedo Jan 29 '12 at 21:36
Complex conjugation is always an element of $Gal(P)$, sometimes it just happens to be trivial (and in the case mentioned, we can be sure it is not). – Tobias Kildetoft Jan 29 '12 at 21:39
@Tobias. If we instead say $P$ has $2k$ non-real roots, then why isn't the statement still true? – Matt Jan 29 '12 at 21:42
@Matt It is. Complex conjugation is in the Galois group of any polynomial over $\mathbb{Q}[X]$. – Ted Jan 29 '12 at 21:44
There is a theorem in many Galois Theory texts where one needs to show there's a transposition in a certain Galois group. If there are exactly 2 non-real roots, then complex conjugation is a transposition. If there are more than 2, complex conjugation won't be a transposition. I wonder if this is what's really behind Matt's questions. – Gerry Myerson Jan 29 '12 at 22:40

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.
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.