If it were in Q then the polynomial would be p(x) = x - π^2, but I don't know if being over Q(π^3) changes that. I think that when you compare the fields you get degree 1, but again, I'm not completely sure.
If it were in Q then the polynomial would be p(x) = x - π^2, but I don't know if being over Q(π^3) changes that. I think that when you compare the fields you get degree 1, but again, I'm not completely sure.
The minimal polynomial here is the monic polynomial $\displaystyle f(x)$ of minimal degree with coefficients in $\displaystyle \mathbb Q\left(\pi^3\right)$ such that $\displaystyle f\left(\pi^2\right)=0.$ If $\displaystyle f(x)$ has degree $\displaystyle 1$ or $\displaystyle 2,$ $\displaystyle f\left(\pi^2\right)$ is of the form $\displaystyle a_0+\pi^2$ or $\displaystyle a_0+a_1\pi^2+\pi^4,$ where $\displaystyle a_0,a_1\in\mathbb Q\left(\pi^3\right);$ clearly these cannot be $\displaystyle 0.$