minimal polynomial of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ I have to find the minimal polynomial of of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ in $\mathbb{Q}$. First I found the minimal polynomial of $\alpha$: $X^5-1=(X-1)(X^4+X^3+X^2+X+1)$ so $X^4+X^3+X^2+X+1$ must be the minimal polynomial of $\alpha$ since its irreducible and has $\alpha$ as a root. I didn't come much further than this. I really need help. Thanks.

Let $\beta=\alpha+\alpha^2$.
Then 
$$ \beta^2=\alpha^4+2\alpha^3+\alpha^2=\alpha^3-\alpha-1$$
$$ \beta^3=\alpha^6+3\alpha^5+3\alpha^4+\alpha^3=3\alpha^4+\alpha^3+\alpha+3=-2\alpha^3-3\alpha^2-2\alpha$$
$$ \beta^4=\alpha^8+4\alpha^7+6\alpha^6+4\alpha^5+\alpha^4=\alpha^4+\alpha^3+4\alpha^2+6\alpha+4=3\alpha^2+5\alpha+3$$
and of course $\beta^0=1$. This gives us five vectors expressed as linear combinations of $1,\alpha,\alpha^2,\alpha^3$, which are therefore linearly dependent. That is you can find nontrivial coefficients $a_4,a_3,a_2,a_1,a_0\in\mathbb Q$ such that $a_4\beta^4+a_3\beta^3+a_2\beta^2+a_1\beta+a_0=0$, that is some nonzero polynomial that has $\beta$ as root.
