# The only algebraic integers in $\mathbb Q$ are the ordinary integers

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

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What are the only rationals that can be a root of $X^n+\ldots +a_1X+ a_0$? This is a result you should know from high-school... –  pki Jan 17 '12 at 17:19
Suppose $\alpha$ is also rational. Then $\alpha = p / q$ for some integers $p$, $q$. Clear denominators... –  Zhen Lin Jan 17 '12 at 17:19
@Paddy The result pki is referring to is the rational root theorem. –  Alex Becker Jan 17 '12 at 17:24
The first example of this one usually learns is that $\sqrt{2}$ is irrational. This is the generalization. –  Michael Hardy Jan 17 '12 at 18:51
What is your definition of an algebraic integer? Usually your lemma is the definition. –  Bill Dubuque Jan 17 '12 at 21:39

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