Let $K \subset L$ be a Galois extension and $x \in L$ such that $L=K[x]$. $H \leq \text{Aut}(L|K)$ is a subgroup of the galois-group.
I want to show that the minimal polynomial of $x$ over $\text{Fix(H)}$ is $$f_x=\prod_{\sigma \in H} (X-\sigma(x)) $$
and in this case if we write $f_x=X^m+a_1 X^{m-1}+...+a_m \in \text{Fix}(H)[X]$ it follows that $\text{Fix}(H)=K[a_1,...a_m]$.
I only know this theorem in the simply case when $\text{Fix}(H)=K[a_1]$. How can I prove this general statement?
You know that the extension $L/L^H$ is Galois and has degree $|H|$ (I assume you know this much Galois theory). Since $L^H[x]=L$ (as $K\subset L^H$), you also have that the minimal polynomial of $x$ over $L^H$ has degree $|H|$. Clearly, $f_x\in L^H[X]$, since the elements of $H$ just permute the factors. Therefore, $x$ is a root of $f_x$ and the degree of $f_x$ coincides with the degree of the minimal polynomial of $x$ over $L^H$.
EDIT: For the other claim, first $K[a_1,\ldots,a_m]\subset L^H$, since the coefficients of $f_x$ are in $L^H$. We also have $K[a_1,\ldots,a_m][x] = K[x] = L$, so that the degree of the minimal polynomial of $x$ over $K[a_1,\ldots,a_m]$ must be $$[L:K[a_1,\ldots,a_m]]\geq [L:L^H].$$ However, by definition, $f_x\in K[a_1,\ldots,a_m][X]$, so the minimal polynomial over $K[a_1,\ldots,a_m]$ must have degree smaller than or equal to $\deg f_x$. Therefore,
$$[L:K[a_1,\ldots,a_m]] \leq \deg f_x = [L:L^H]$$
and so
$$[L:L^H][L^H:K[a_1,\ldots,a_m]]=[L:K[a_1,\ldots,a_m]]= [L:L^H]$$
and we can deduce that $L^H=K[a_1,\ldots,a_m]$.
