Exercise 1.1.6 (Conjugacy of Rational)

Proof. We must understand here "conjugate" as "conjugate over $\mathbb{Q}$".

Let $z=a\in \mathbb{Q}$, and $z^{\prime }=c+\mathit{id}\in ℂ$, with $x,y\in ℝ$. Assume that $z^{\prime }$ is conjugate to $z$ over $\mathbb{Q}$. Consider the polynomial $p(t)=t-a\in \mathbb{Q}[t]$, where $t$ is a variable. Then $p(z)=p(a)=0$. Since $z^{\prime }$ is conjugate to $z$ over $\mathbb{Q}$, $p(z^{\prime })=0$, thus $0=c+\mathit{id}-a=(c-a)+\mathit{id}$, with $c-a\in ℝ,d\in ℝ$. Therefore $d=0$, and $c=a$. This gives $z^{\prime }=c+\mathit{id}=a=z$.

This proves that $z\in \mathbb{Q}$ is not conjugate to $z^{\prime }$ over $\mathbb{Q}$ for any complex number $z^{\prime }\ne z$. □