# Complex roots of an irreducible polynomial

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 $p$?

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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?
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
@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