finding the minimal polynomial Let $K=\mathbb{Q}[\omega]$ where $\omega^2+\omega+1=0$ and let $R$ be the polynomial ring $K[x]$. Let $L$ be the field $K(x)[y]$ where $y$ satisfies $y^3=1+x^2$.
Take an element $\alpha\in L$. How can we find the minimal polynomial of $\alpha$ over $K(x)$?
 A: Since $\omega$ is a primitive cube root of $1$ and $K(x) \subset L$ is a cubic extension, the extension is Galois and the Galois group is obtained by replacing $y$ with $y, \omega y, \omega^2 y$ : concretely, those automorphisms are the three maps $L \to L$ sending $f(y) \mapsto f(y), f(\omega y), f(\omega^2 y)$.
Now, if you have an element $\alpha(y) \in L$, if $\alpha(y) = \alpha(\omega y)$ then $\alpha(y) \in K(x)$ : you can write $\alpha$ as a rational function of $y^3$ (there is a $\beta \in K(x)[Z]$ such that $\alpha(y) = \beta(y^3)$), and then $\alpha(y) = \beta(1+x^2) \in K(x)$, so its minimal polynomial is $Z - \beta(1+x^2)$.
If $\alpha(y) \neq \alpha( \omega y)$, then $\alpha(y) \notin K(x)$, so $\alpha(y)$ has $3$ distinct conjugates, and its minimal polynomial is $(Z - \alpha(y))(Z- \alpha(\omega y))(Z - \alpha(\omega^2 y))$. If you develop this polynomial, you will find that you can express each coefficient as a rational function of $y^3$ (just like in the previous case), which means that they are in $K(x)$ as needed.
