Minimal polynomial for some galois extension The following problem is an exercise in Dummit & Foote's textbook - Abstract Algebra(3rd) in section 14.4:

Suppose that $K/F$ is Galois with Galois group $G$, and $\theta$ is a primitive element for $K$, i.e., $K=F(\theta)$. For any subgroup $H$ of $G$, let $$f(x):=\prod_{\sigma\in H}\big(x-\sigma(\theta)\big).$$

*

*Show that $f(x)\in E[x]$ where $E$ is the fixed field of $H$ in $K$, and that $f(x)$ is the minimal polynomial for $\theta$ over $E$.


*Prove that the coefficients of $f(x)$ generate $E$ over $F$.

In view of this problem, I wonder if it can be applied to solve the following problem:

Let $K=\mathbb{Q}(\alpha)$ be a simple extension over $\mathbb{Q}$(possibly non-Galois over $\mathbb{Q}$) where $\alpha$ is algebraic over $\mathbb{Q}$, and let $F$ be a subfield of $K$. Suppose that the minimal polynomial for $\alpha$ over $F$ given by $$\textrm{irr}(\alpha,F):=x^{r}+a_{1}x^{r-1}+\cdots+a_{r-1}x+a_{r},$$ where $a_{1},a_{2},\ldots,a_{r}\in F$.  Show that $F=\mathbb{Q}(a_{1},a_{2},\ldots,a_{n})$.

When I think as follows, I thought that the first and second problems have the same meaning:

If we consider the Galois closure $L$ of $K$ over $F$, and take the fixed field $L_{H}$ of $H:=\textrm{Gal}(L/F)$ in $L$, then we can prove that $$f(x)=\prod_{\sigma\in H}\big(x-\sigma(\alpha)\big)$$ is exactly the minimal polynomial for $\alpha$ over $L_{H}$, and which derive $F=\mathbb{Q}(a_{1},a_{2},\ldots,a_{n})$.

It seems possible, but it is difficult to fill in the details.
Can anyone help me a little? Thank you.
 A: I will give a proof for your first problem, which will also 
prove the second statement. The point is that the proof of the second part 
of first problem doesn't require $K/F$ being Galois 
(only simple extension is enough) so it can be 
copied to the second problem. 
Proof of first part. Since $E\subset K$ is fixed field of subgroup $H\subset \text{Aut}(K)$ so $K/E$ is a Galois 
extension with Galois group $H$. This follows $|H|=[K:E]$.
Let $f\in E[x]$ be minimal polynomial of $\theta$ over $E$ then 
as $K=F(\theta)=E(\theta)$ so $\text{deg}(f)=[K:E]=|H|$.
On the other hand, as $H=\text{Aut}(K/E)$ so $\sigma(\theta)$ 
is root of $f$ for all $\sigma\in H$. This follows $f(x)=\prod_{\sigma\in H}
(x-\sigma(\theta))$.
Proof of second part. Let $f(x)=x^r+a_{r-1}x^{r-1}+\cdots+a_1x_1+a_0$
then $a_i\in E$ so $F(a_0,\ldots, a_{r-1})\subset E\subset K$. 
This inclusion implies two things:
One is that $[K:F(a_0,\ldots, a_{r-1})]\ge [K:E]$.
Second,
the minimal polynomial $g$ of $\theta$ over $F(a_0,a_1,\ldots,a_{r-1})$ 
must divide $f$, implying $\deg g\le \deg f=[K:E]$. 
On the other hand, since $K=F(\theta)=F(a_0,\ldots, 
a_{r-1})(\theta)$ is 
simple so $\deg g=[K:F(a_0,\ldots, a_{r-1})]$. 
Thus, $[K:F(a_0,\ldots, a_{r-1})]=[K:E]$ so $E=F(a_0,\ldots, a_{r-1})$, 
as desired. 
