Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a part of a question from a Berkeley prelim exam and I would appreciate a hint, since I can't see a promising approach to this.

Let $p$ be an irreducible polynomial over the rationals with a nonzero complex root $a$. Suppose that $a^2$ is also a root of $p$. How would one conclude from this that for any root $b$ of $p$, $b^2$ is also a root of $p$?

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Hint: The number $a$ is a zero of the polynomials $p(x)$ and $q(x)=p(x^2)$. Show that this means that $a$ is also a root of the greatest common divisor $h(x)=gcd(p(x),q(x))\in\mathbf{Q}[x]$. What can you say about the polynomial $h(x)$ using the fact that $p(x)$ was known to be irreducible?

share|improve this answer
    
This is easier, if you are familiar with the concept of a minimal polynomial of an algebraic number. –  Jyrki Lahtonen Nov 3 '11 at 10:12
    
Right. As $\mathbb{Q}[x]$ is a PID, a variant of Bezout's lemma holds and $a$ is a root of $h(x)$. Since $p$ was irreducible and $h$ cannot be constant, as it has a root, we have $h=cp$ for some nonzero constant $c$. Therefore $p$ divides $q$ and we're done. –  Miha Habič Nov 3 '11 at 11:01
    
I wonder what you meant by your comment on the minimal polynomial. I am familiar with the material, but I'm not sure how to use it here. –  Miha Habič Nov 3 '11 at 11:02
    
@MihaHabič: Correct. Instead of Bezout, you can also simply look at Euclid's algorithm computing the polynomial gcd. Explaining the comment: $p(x)$ is the minimal polynomial of $a$ over $\mathbf{Q}$, so automatically any polynomial with rational coefficients that has $a$ as a zero is a multiple of $p(x)$. Here we apply this to $q(x)=p(x^2)$. –  Jyrki Lahtonen Nov 3 '11 at 11:13
    
Or in another way: $p$ is the minimal polynomial of $a$ over $\mathbb{Q}$. Hence we have $\mathbb{Q}(a) \cong \mathbb{Q}\left\[x \right\] /(p) \cong \mathbb{Q}(b)$ for every root $b$ of $p$ via $a \mapsto \overline{x} \mapsto b$ and since $a^2$ is a root of $p$ we get $\overline{x}^2$ is a root of $p$ and therefore $b^2$ is a root of $p$ for every root $b$ of $p$. –  Sebastian Schoennenbeck Nov 3 '11 at 13:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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