$[F(a):F(a^i)](y_1+...+y_m)=x_1^i+...+x_n^i$

Question

$E=F(a)$, $E$ is a finite dimensional separable extension of $F$. If $x_1,...,x_n$ are roots of minimal polynomial of $a$ over $F$ is it true that for an arbitrary $i$, $[F(a):F(a^i)](y_1+...+y_m)=x_1^i+...+x_n^i$ where $y_1,...y_m$ are roots of minimal polynomial for $a^i$ over $F$?

Answer 1

For $F(a)/F$ a finite separable extension, let $f_i\in F[t]$ be $a^i$'s minimal polynomial, let $L$ be $f_1$'s splitting field, let $Hom_F(E,L)$ be the set of $F$-algebra embeddings $E\to L$, then $$\sum_{b\in L,f_i(b)=0} b=\sum_{g\in Hom_F(F(a^i),L)} g(a^i)$$ $$ =\sum_{g\in Hom_F(F(a),L)}\frac{ g(a^i)}{|Hom_{F(g(a)^i)}(F(g(a)),L)|}=\sum_{g\in Hom_F(F(a),L)}\frac{ g(a)^i}{[F(g(a)):F(g(a)^i)]}$$ $$ =\sum_{g\in Hom_F(F(a),L)}\frac{ g(a)^i}{[F(a):F(a^i)]}=\frac1{[F(a):F(a^i)]}\sum_{c\in L,f_1(c)=0} c^i$$

Note this proof works the same way when replacing $a^i$ by $p(a)$ for some $p\in F[t]$.