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If the rational number $r$ is also an algebraic integer, show that $r$ is an integer

So $r$ is a rational number that is a root of a polynomial $p(x)$ over the integers, right? I need to show that $r$ is an integer.

I know that for $p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0$

we have that $a_0$ is the product of all roots. I don't think it helps because we could have two roots $a/b$ and $b/a$ and their product is an integer, so the fact that the polynomial is over the integers won't do anything in this case.

Somebody have an idea?

Guerlando OCs
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    This is classic. Hint: write $r$ in lowest terms, i.e., $r=p/q$ such that $\gcd(p,q)=1$. – Xam Sep 19 '17 at 05:30
  • Do you know the rational roots theorem? – TomGrubb Sep 19 '17 at 05:31
  • The comment made by Gerry Myerson (or Arturo's answer) in the other question should be enough to give you a clue about how to prove this result. And yes, yours is a particular case. – Xam Sep 19 '17 at 05:40

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