Please help me with the following question.
Let $f(x)$ be n-th degree irreducible polynomial in $\mathbb Q[x].$ And let $a_i$ for $i\in [1,n]$ be the roots of $f(x).$ and for each $a_i$ let $d_i(f)$ be the roots of $f(x)$ in $\mathbb Q(a_i)$. Then show that $d_1(f)=...=d_n(f).$
Now let $d(f):=d_1(f)=...=d_n(f).$ Then show that $d(f)$ is the number of automorphisms for the field $\mathbb Q(a)$ where $a$ is a root of $f(x)$.
Since is irreducible it means we can define an isomorphism by and for . Now for it means . Let , from here we see that is an isomorphism with and for .
Consider . The order of this Galois group is equal to the number of roots of in i.e. . Therefore, to show that it sufficies to prove that . To prove this notice that if then . This means, . By a mirror argument . It follows that .