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?


  • 5
    $\begingroup$ 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 '12 at 17:19
  • 3
    $\begingroup$ Suppose $\alpha$ is also rational. Then $\alpha = p / q$ for some integers $p$, $q$. Clear denominators... $\endgroup$ – Zhen Lin Jan 17 '12 at 17:19
  • 5
    $\begingroup$ @Paddy The result pki is referring to is the rational root theorem. $\endgroup$ – Alex Becker Jan 17 '12 at 17:24
  • 1
    $\begingroup$ The first example of this one usually learns is that $\sqrt{2}$ is irrational. This is the generalization. $\endgroup$ – Michael Hardy Jan 17 '12 at 18:51
  • 2
    $\begingroup$ What is your definition of an algebraic integer? Usually your lemma is the definition. $\endgroup$ – 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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.