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
Best Answer
Say $f(X)=X^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$ and $\alpha=p/q$ is a root in simplest form. Then
$$q^nf(p/q)=p^n+a_{n-1}p^{n-1}q+\cdots+a_1pq^{n-1}+a_0q^n=0.$$
Reduce both sides modulo $q$ and invoke unique factorization (the fundamental theorem of arithmetic) to derive a contradiction (this is unless $q=1$, of course).