## A fact about divisibility in a splitting field

Let $E = \mathbb{Q}(\theta)$ be an algebraic extension of $\mathbb{Q}$, and let $K = \mathbb{Q}(\theta_1,\ldots,\theta_k)$, where the $\theta_i$ are the conjugates of $\theta$. Let $\alpha,\beta \in E$, and denote by $\alpha_i$ and $\beta_i$ the $i$th conjugate of $\alpha$ and $\beta$ (respectively). Show that if $\beta|\alpha$ in $E$, then $\beta_i|\alpha_i$ in $E$.

Suppose $\alpha/\beta$ is an algebraic integer in $K$. In particular, $\alpha/\beta \in F(\theta)$, so that $\alpha/\beta = r(\theta)$ for some polynomial $r(x)$.

Now $\alpha/\beta$ is a root of some irreducible monic polynomial $h(x)$ with rational integer coefficients. That is, $h(r(\theta)) = 0$. In particular, $\theta$ is a root of $h \circ r$, so that the conjugates $\theta_i$ are also roots of $h \circ r$. So $h(r(\theta_i)) = 0$ for each $\theta_i$, and thus $\alpha_i/\beta_i = r(\theta_i)$ is a root of a monic polynomial with rational integer coefficients. So $\alpha_i/\beta_i$ is an algebraic integer in $K$ (in fact in $\mathbb{Q}(\theta_i)$).