Let $\alpha\in\Bbb{C}$ algebraic, root of $X^5+2X+6$. The question is exist a $\Bbb{Q}$-isomorphism $\sigma:\Bbb{Q}(\alpha) \to \Bbb{Q}(\alpha)$ such that $\sigma(\alpha)=\alpha ^{-1}$?
I know this is false (I cheked its roots in Maple), and I know that it is true only if $\alpha ^{-1}$ is root of the polynomial too, but how I solve this in an exam? I couldn't find the roots. How do you solve this type of exercises?
Thanks for your time.
Let $f(X)=X^5+2X+6$ and note that $f$ is the minimal polynomial for $\alpha$ since $f$ is irreducible (it's 2-Eisenstein). If $\sigma(\alpha)=1/\alpha$ then $f(1/\alpha)=0$ as well. Hence $$ \alpha^5f(1/\alpha)=1+2\alpha^4+6\alpha^5=0, $$ so you now have $\alpha^5=-(1+2\alpha^4)/6$. But you also know, since $f(\alpha)=0$, that $\alpha^5=-(2\alpha+6)$. Equating these, you can find a degree 4 polynomial whose root is $\alpha$, contradicting the fact that $\alpha$ has degree $5$ over $\mathbb{Q}$.