# Closure of Algebraic Field to Complex Conjugation

I have an algebraic field $\mathbb Q(\gamma)$ with $\gamma$ the complex root of $X^3+X^2+X-1$, i.e., $\gamma\approx-0.771+1.115\mathrm i$.

I have two closely related questions:

1. Is $\mathbb Q(\gamma)$ closed under complex conjugation?

2. If so, how can one express $\overline\gamma$ as an element of $\mathbb Q(\gamma)$, i.e. as $\overline\gamma=a+b\gamma+c\gamma^2$ for $a,b,c\in\mathbb Q$?

And it is possible to answer this question more in general?

In your example, ${\bf Q}(\gamma)$ is not closed under complex conjugation. There may be another way to do this, but I look at it using Galois Theory. The extension $K$ of the rationals generated by the full set of roots has Galois group a subgroup of $S_3$ (because the polynomial has degree 3), of order a multiple of 3 (ditto), and a multiple of 2 (since the group contains complex conjugation), so the group is all of $S_3$. Thus $K$ has degree 6 over the rationals (since that's the order of the Galois group); if $\overline\gamma$ were in ${\bf Q}(\gamma)$, then $K$ would be ${\bf Q}(\gamma)$, and would have degree only 3 over the rationals.
This argument works for all irreducible cubics over the rationals; if there is a nonreal root $\gamma$, then $\overline\gamma$ is not in ${\bf Q}(\gamma)$.
• I'm not sure I follow the part: " if $\;\overline\gamma\in\Bbb Q(\gamma)\;$ then $\;K=\Bbb Q(\gamma)\;$ "...why? We still would be missing the real root that, for sure, the cubic has, wouldn't we? – DonAntonio Nov 18 '13 at 12:54
• @Don, call the roots $r,s,t$; they add to $-1$, so if two of them are in a field, so is the third. – Gerry Myerson Nov 18 '13 at 12:56