I'm trying to determine the minimal polynomials of $-\alpha$, $1-\alpha$, $2\alpha$, and $1/\alpha$, given that the minimal polynomial of $\alpha$ is $x^3-x-1$.
Since $x^3-x-1$ is the minimal polynomial of $\alpha$, we have $\alpha^3-\alpha-1=0$. So, $\alpha^3=\alpha+1$, and $-(\alpha^3-\alpha-1)=0$, so $-\alpha^3+\alpha+1=0$, and thus $-\alpha$ satisfies $x^3+x+1=0$. Would that be the minimal polynomial? How would I get the other minimal polynomials requested? Thanks!
you can solve in this way: let $\beta=f(\alpha)$ write $\alpha=g(\beta)$ and put g into the minimal polinomial, the result would be a function of $\beta$ if it is not a polinomial then transform it, this is the minimal polinomial you look for. For example $\beta=1-\alpha$ so $\alpha=1-\beta$ and $(1-\beta)^3-(1-\beta)-1=0$ you just have to solve it.