Lorenzini's "Invitation to arithmetic geometry", 1st exercise

If $\alpha\in\Bbb C$ and $\Bbb Z[\alpha]$ is a finitely generated Abelian group, then $\alpha$ is an algebraic integer.

To see this, write $\Bbb Z[\alpha]=\Bbb Z\beta_1+\cdots+\Bbb Z\beta_n$. Then $\alpha\beta_j\in\Bbb Z[\alpha]$ so that $\alpha\beta_j=\sum_{k=1}^n c_{jk}\beta_k$ with the $c_{jk}\in\Bbb Z$. Then $\alpha$ is an eigenvalue of the integer matrix $C=(c_{jk})$, and so is an algebraic integer.

For $d\in\Bbb Q$, $\sqrt d$ is only an algebraic integer when $d\in\Bbb Z$.