How to compute the irreducible polynomial of a number

Question

Set $\alpha \in \mathbb{R}$ and $f=x^3+5x+1$ such that $f(\alpha)=0$. How could I compute the irreducible polynomial over $\mathbb{Q}$ of $\alpha ^2$? And of $\sqrt{3}\alpha$ ?

The problem is I am out of ideas since I don't know explicity $\alpha$.

Answer 1

We know that $\alpha^3+5\alpha+1=0$, so multiplying by $\alpha$ gives (letting $\alpha^2=b$) $b(b+5)=-\alpha$; squaring gives $b^2(b+5)^2=b$ or $b^3+10b^2+25b-1=0$. This is the minimal polynomial for $b$.

Now return to $f$ and multiply by $\sqrt3$ on both sides, giving (letting $\sqrt3\alpha=c$) $c^3/3+5c=-\sqrt3$. Again, squaring gives $c^6/9+10c^4/3+25c^2-3=0$, the minimal polynomial for $c$.

Answer 2

Hint: Let $\beta = \alpha^2$. Find the matrix of the map $x \mapsto \beta x$ in the basis $\{1,\alpha,\alpha^2\}$.