Minimal polynomial of $\theta^2$ I'm doing this Galois Theory question:
$f = x^3 + x + 3$ is known to be irreducible and has just one real root, call it $\theta$. 
What is the minimal polynomial for $\theta^2$? Is it just $f$? 
 A: The minimal polynomial of $\theta^2$ must have degree at most $3$.
So, we need to look for a linear equation for $1, \theta^2, \theta^4, \theta^6$.
$\theta^3 = -\theta-3$ implies $\theta^4= -\theta^2-3\theta$ and $\theta^6= \theta^2+6\theta+9$. So $\theta^6+2\theta^4=-\theta^2+9$.
A systematic way of refining this relation is to solve a linear system $Ax=0$, where the columns of $A$ are the coordinates of $1, \theta^2, \theta^4, \theta^6$ in the basis $1, \theta, \theta^2$.
A: There’s another systematic way of doing this, involving the Norm. Let $R=\Bbb Q[X]$, and look at $S=R[\theta]/(\theta^3+\theta+3)$, and the Norm from $S$ down to $R$. You can calculate the Norm of an element $z$ by looking at the $R$-linear map $w\mapsto zw$, and calculating its determinant. It’s just a $3\times3$ with coefficients in $R$, not bad at all to do.
But the Norm of $X-\theta^2$ is clearly a cubic polynomial in $R=\Bbb Q[X]$ that has $\theta^2$ for a root. In this case, it is irreducible, and the computation gives $-9 + X + 2X^2 + X^3$.
