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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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    $\begingroup$ 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... $\endgroup$
    – pki
    Jan 17, 2012 at 17:19
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    $\begingroup$ Suppose $\alpha$ is also rational. Then $\alpha = p / q$ for some integers $p$, $q$. Clear denominators... $\endgroup$
    – Zhen Lin
    Jan 17, 2012 at 17:19
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    $\begingroup$ @Paddy The result pki is referring to is the rational root theorem. $\endgroup$
    – Alex Becker
    Jan 17, 2012 at 17:24
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    $\begingroup$ The first example of this one usually learns is that $\sqrt{2}$ is irrational. This is the generalization. $\endgroup$
    – Michael Hardy
    Jan 17, 2012 at 18:51
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    $\begingroup$ What is your definition of an algebraic integer? Usually your lemma is the definition. $\endgroup$
    – Bill Dubuque
    Jan 17, 2012 at 21:39

2 Answers 2

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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).

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

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