If $\alpha$ is algebraic over a Field, find a nonzero polynomial such that $q(\alpha^2) = 0$

Question

Let $\alpha$ be algebraic over a field, $F$, with minimal polynomial $p(t) = t^3+t+1$. Find a nonzero polynomial $q(t) \in F[t]$ such that $q(\alpha^2) = 0$.

Now, I get that $\alpha^3 = -\alpha - 1$ and that $\alpha^4 = -\alpha^2-\alpha$ and that $\alpha^5 = -\alpha^2 + \alpha + 1$ and so on - but I don't know how to use this information to get the answer?

Answer 1

We know that $\alpha^{3}+\alpha+1=0$, hence also $$(\alpha^{3}+\alpha+1)^{2}=\alpha^{6}+2\alpha^{4}+2\alpha^{3}+\alpha^{2}+2\alpha+1=0.$$ As $2\alpha^{3}+2\alpha+1=2(\alpha^{3}+\alpha+1)-1=-1$, this implies that $$\alpha^{6}+2\alpha^{4}+\alpha^{2}-1=0.$$ So $\alpha^{2}$ is a root of the polynomial $q(t)=t^{3}+2t^{2}+t-1$.

Answer 2

Gather the even and odd powers of $t$ in the minumum polynomial of $\alpha$: $$\alpha^3 + \alpha = 1.$$ Now square both sides to obtain $$\alpha^6 + 2 \alpha^4 + \alpha^2 = 1.$$ Therefore, $\alpha^2$ is a root of $q(t) = t^3 + 2 t^2 + t - 1$.