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.

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.)
