How to finds the minimal polynomial of $α + α ^2$ over $\Bbb Q$.?

Question

Let $p$ be a prime and $K = \Bbb Q(α)$, where $α ^3 = p$. Find the minimal polynomial of $α + α ^2$ over $\Bbb Q$.

my attempt : I was taking $α + α ^2$= $α(1 + α )$=$0$ and getting $α =0$ and $α= -1$ as I don't know how to proceed further

Please, Help me.

Answer 1

Hint: start with $x=\alpha + \alpha^2\Rightarrow x-\alpha^2=\alpha$ then try to cube it and then try to make the coefficients of the polynomial in $\Bbb Q$.

Adding details:

After taking cube $x-\alpha^2=\alpha$ we get $x^3-3\alpha^2x^2+3\alpha^4 x-\alpha^6=\alpha^3=b \Rightarrow x^3-3\alpha^2x^2+3p\alpha x-p^2=b$

Now observe that $x^3-p=0\Rightarrow \big(\frac x {\alpha}\big)^3-1=0\Rightarrow \alpha^2+x\alpha+x^2=0$. Now you take from here.

Answer 2

$$A=\pmatrix{0&1&0\\0&0&1\\p&0&0}$$ has $\alpha$ as an eigenvalue (for eigenvector $\pmatrix{1\\\alpha\\\alpha^2}$). Then $B=A+A^2$ has $\alpha+\alpha^2$ as an eigenvalue, so that $f(\alpha)=0$ where $f$ is the characteristic polynomial of $B$.

Answer 3

A fully automated way to approach such questions is to use resultants or Gröbner bases. For example in this case the resultant of $x^3-p$ and $y - (x^2+x)$ with respect to $x$ turns out to be $y^3 - 3 p y - p^2 - p$ which happens to be irreducible. The same polynomial appears in the Gröbner basis of the ideal $(x^3-p, y-(x^2+x))$ with respect to the lexicographical order with $x>y>p$.