# Minimal polynomial of $\alpha^2$ given the minimal polynomial of $\alpha$

Given that $\alpha$ is a root (in the field extension) of the irreducible polynomial $X^4+X^3-X+2\in\mathbb{Q}[X]$, I have to find the minimal polynomial of $\alpha^2$. I am thinking about this for a while, but I can't find it. I need some hints. Thank you.

You have $\alpha^4 + 2 = \alpha - \alpha^3$, and squaring both sides gives a polynomial of degree $4$ satisfied by $\alpha^2$. (All the powers of $\alpha$ appearing will be even.) Then show it is the minimal polynomial for $\alpha^2$.

• Nice; so that's the clever way. – Magdiragdag Dec 17 '13 at 19:31

Maybe someone can come up with some clever way, but here is the hard-work method.

Write down $1, \alpha^2, \alpha^4, \alpha^6, \alpha^8$ in terms of $1, \alpha, \alpha^2, \alpha^3$. These five elements should be linearly dependent, so you'll find $c_0 1 + c_1 \alpha^2 + c_2 \alpha^4 + c_3 \alpha^6 + c_8 \alpha^8 = 0$ for some $c_i \in {\mathbb Q}$. This gives you a 4th degree expression in $\alpha^2$ that equals 0.

(Note, if the degree of the extension ${\mathbb Q}(\alpha^2) : {\mathbb Q}$ were 2, you'd already find $1, \alpha^2, \alpha^4$ to be linearly dependent, but you'll see directly that they're not. The degree can't be 3, because it must be a divisor of $[{\mathbb Q}(\alpha) : {\mathbb Q}] = 4$. So it must be 4.)

Look up the Dandelin-Graeffe iteration. In short, $p(x)\cdot p(-x)$ is a polynomial in $x^2$, say $q(x^2)=p(x)\cdot p(-x)$. So if $p(\alpha)=0$, then $q(\alpha^2)=0$.

This gives the same result as the answer by Zarrax, however $q$ need not be irreducible.