TITLE: The only algebraic integers in $\mathbb Q $ are the ordinary integers
QUESTION [8 upvotes]: I think I'm being a bit slow here.
Lemma: Every algebraic integer is the root of some monic irreducible polynomial with coefficients in $\mathbb Z$. 
Corollary: The only algebraic integers in $\mathbb Q$ are the ordinary integers.
I'm struggling to see how this corollary follows from the lemma. Suppose $\alpha$ is an algebraic integer. Then there is a monic, irreducible $f$ with integer coefficients such that $f(\alpha) = 0$. Why can't $\alpha$ be a non-integer?
Thanks

REPLY [2 votes]: Then $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$ since $\alpha \in \mathbb{Q}$, so the minimal polynomial has degree 1. You're done.