I am trying to understand the proof of the following statement:
Let $\mathscr{θ}$ be an algebraic element over the finite field $F$ and $\mathscr{θ=θ_1,θ_2 ... θ_n}$ be all the conjugate elementes of $\mathscr{θ}$ over $F$. Also let $s(x)\in F[x]$ and $\mathscr{α}=s(\mathscr{θ})$ and $\mathscr{β}$ is conjugate element of $\mathscr{α}$ over $F$. Then $\mathscr{β}=s(θ_i)$ for some $1\leq i \leq n$.
Here is what the proof is:
Consider the polynomial $g(x)$ = $\prod_{i=1}^{n}(x-s(θ_i))$. This polynomial is monic and $g(x)\in F[x]$. Then let $f_\mathscr{α}\in F[x]$ be the minimal polynomial of $\mathscr{α}$. It follows from $g(\mathscr{α})=0$ that $f_\mathscr{α}(x)|g(x)$. Then all conjugate elements of $\mathscr{α}$ over $F$ (including $\mathscr{β}$) are amongst the roots of $g(x)$ which means they are $s(θ_i)$.
I don't know how to prove that $g(x)\in F[x]$. Actually there is a hint how to prove it - I should use a technique similar to the one used in another proof. And here is my atempt :
Let $K=F(\mathscr{θ_1,θ_2 ... θ_n})$ and $\mathscr{θ_1,θ_2 ... θ_n}$ are roots of $f(x)\in F[x]$. Next I need $f(x)=\prod_{i=1}^{n}(x-θ_i)$. If we assume that this is true then according to Viète theorem the coefficients of $f(x)$ are the elementary symmetric polynomials of the roots $\mathscr{θ_1,θ_2 ... θ_n}$ which means that the coefficients of $f(x)$ are elements of the field $F$. $g(x)$ = $\prod_{i=1}^{n}(x-s(θ_i))$ in other hand again according to Viète theorem has coefficients which are the elementary symmetric polynomials of the roots $s(θ_1),s(θ_2) ... s(θ_n)$. Each coefficient of $g(x)$ should be a symmetrical polynomial again and then I can use the fundamental theorem of symmetric polynomials to prove that $g(x)\in F[x]$ but I can't find a solution. Perhaps this approach is wrong.
Any suggestions and explanations will be very much welcome! Thank you!