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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

  • 5
    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...2012-01-17
  • 3
    Suppose $\alpha$ is also _rational_. Then $\alpha = p / q$ for some integers $p$, $q$. Clear denominators...2012-01-17
  • 4
    @Paddy The result pki is referring to is the [rational root theorem](http://en.wikipedia.org/wiki/Rational_roots_theorem).2012-01-17
  • 1
    The first example of this one usually learns is that $\sqrt{2}$ is irrational. This is the generalization.2012-01-17
  • 2
    What is your definition of an algebraic integer? Usually your lemma is the definition.2012-01-17

2 Answers 2