substituting algebraic numbers for their conjugates in rational functions

Question

If I have two algebraic numbers $\alpha$ and $\beta$ and a rational function $w$ with rational coefficients (a function that's the ratio of two rational polynomials) that relates the two $\alpha = w(\beta)$ if I were to substitute $\beta$ with one of it's algebraic cojugates $\beta^{'}$ in the rational function will I get a conjugate of $\alpha$ or rather is $w(\beta^{'})$ an algebraic conjugate of alpha? I feel like there is a very obvious counter example but I've been struggling to find one.

Answer 1

Let $f,F\in\Bbb Z[X]$ be irreducible. Let $g,h\in \Bbb Z[X]$ with $h\ne 0$. Let $\beta\in\Bbb C$ with $f(\beta)=0$ and $F(\frac{g(\beta)}{h(\beta)})=0$ (and in particular, $h(\beta)\ne 0$). Note that by cancellation of $h$'s, $$p(X):=h(X)^{\deg F}F(\tfrac{g(X)}{h(X)})$$ is actually $\in \Bbb Z[X]$. As $p(\beta)=0$, we conclude that $\beta$'s minimal polynomial $f$ divides $p$. On the other hand, $h(\beta)\ne 0$ implies that $f$ does not divide (or even has a common factor with) $p$. But that means that for every $\beta'$ with $f(\beta')=0$, we have $p(\beta')=0$ and $h(\beta')\ne 0$, hence $$F(\tfrac{g(\beta')}{h(\beta')})=0,$$ i.e., $\tfrac{g(\beta')}{h(\beta')}$ is a conjugate of $\tfrac{g(\beta)}{h(\beta)}$.