## Divisibility among rational integers in an algebraic number field

Let $E = \mathbb{Q}(\theta)$ be an algebraic extension of $\mathbb{Q}$ and let $a,b \in \mathbb{Z}$. Recall that if $\alpha,\beta \in E$ are algebraic integers, we say that $\beta|\alpha$ if the quotient $\alpha/\beta$ is an algebraic integer. Prove that $b|a$ in $E$ if and only if $b|a$ in $\mathbb{Q}$.

If $b|a$ in $\mathbb{Q}$, then $a/b$ is a rational integer. Certainly $a/b \in E$ is an algebraic integer; so $b|a$ in $E$.

Now suppose $b|a$ in $E$. Since $a/b$ is an algebraic integer and a rational number, $a/b$ is a rational integer. So $b|a$ in $\mathbb{Q}$.