# Constructible and Rational Roots

Prove a cubic polynomial with rational coefficients $$f$$ does not have a root in $$Q$$ implies $$f$$ does not have a constructible root in the reals.

My attempt at the proof goes as follows: If we suppose that f does not have a root in $$Q$$, then $$f$$ either has complex root(s) or irrational roots. More generally, we can say $$f$$ has a repeated real roots, $$f$$ has two complex conjugate roots and one real root, of $$f$$ has three distinct roots. What I'm having trouble showing is how to eliminate these cases. That is, I'm unable to find a way to show that one of these cases will lead to a violation of $$f$$ having integer coefficients. And if that is violated, then I can follow up by stating that there does not exist a chain of field extensions which would culminate with showing that $$f$$ does not have a constructible roots in the reals. Any help would be appreciated.

• Please use MathJax and line breaks to format your posts. – jgon Mar 25 at 22:21
• I made the suggestion because you will find more people willing to help you with your problem the easier you make it for them to understand your problem. – jgon Mar 25 at 22:47

If $$f$$ doesn't have a rational root, then $$f$$ is an irreducible cubic in $$\Bbb{Q}[x]$$. As a result, its roots all have degree $$3$$ over $$\Bbb{Q}$$, so they aren't constructible.
• @SanjoyTheManjoy The reason is that if $f=gh$, with $g$ and $h$ nonconstant and $f$ of degree 3, then one of $g$ or $h$ has degree one, i.e., one of the factors of $f$ is linear when it isn't irreducible. Thus when $f$ is not irreducible, it has a root in the base field. – jgon Mar 27 at 2:31