Normal extension and group of automorphisms Let $K \supset F$ a normal extension (finite). Show in details that, for every $\alpha \in K$, the polynomial
$$f(x)=\prod_{\sigma \in Aut_F(K)}(x-\sigma(\alpha))\in F[x];$$
My attempt:
Since $K$ is a finite extension of $F$, $K$ is an algebraic extension, so $\alpha$ is algebraic over $F$.
If $\alpha \in F$, then $\sigma(\alpha)=\alpha$ for every $\sigma \in$ Aut$_F(K)$, and, obviously, $f(x)=(x-\alpha)^k \in F[x]$, where $k$ $=$ |Aut$_F(K)$|.
If $\alpha \notin F$, we know that $\alpha$ is algebraic over $F$, so there is $p(x)=irr(\alpha,F) \in F$. Since, the extension is normal, every root of $p(x)$ is in $K$ and, it's know that any automorphism in Aut$_F(K)$ will take a root of an irreducible polynomial into another, so if $\{\alpha=\alpha_1,\ldots,\alpha_m\}$ is the set of rotos of $p(x)$, any automorphism will permute these roots. 
We can write $p(x)=(x-\alpha_1)^{n_1}\ldots(x-\alpha_m)^{n_m} \in K[x]$.
I can't go any further than that... My initial idea was to show that $f(x)=p(x)$, however I see that it's not true, because the number of automorphisms in Aut$_F(K)$ can be bigger than $p$'s degree, so I can't see how to guarantee that $f(x) \in F[x]$.
I know that $f(x)=(x-\alpha_1)^{n'_1}\ldots(x-\alpha_m)^{n'_m}$ , $0\leq n'_i\leq k=$|Aut$_F(K)$|. So if $n'_i \geq n_i$, we will have that $f(x)=p(x)g(x)$, $g(x)=(x-\alpha_1)^{n'_1-n_1}\ldots(x-\alpha_m)^{n'_m-n_m}$. So I'd have to show that $g(x)\in F[x]$, but I can't see how to do that... And, even if I knew, it doesn't solve the entire problem.. 
Does someone have a better idea to solve this?
Thanks :)
 A: The necessary datum is the following:

Lemma Let $L/k$ be a normal and separable extension. Then the intersection of the fixed fields of automorphisms of $L/k$ is exactly $k$, that is, $k=\{x\in L:\sigma x=x\forall\sigma\in\text{Aut}(L/k)\}$.

From this we simply point out that $f$ is fixed by automorphisms of $L$, so that all of its coefficients are as well, and they're thus in $k$.
Proof of lemma Let $k$ and $L$ be embedded in a common algebraic closure $\bar k$. By definition, all automorphisms of $L$ over $k$ fix $k$. Then let $\alpha\in L\setminus k$, and $\bar\alpha\neq \alpha$ be another root of $\alpha$'s minimal polynomial. Then $k(\alpha)$ is isomorphic to $k(\bar\alpha)$ by $\alpha\mapsto \bar\alpha$. This automorphism extends to $\bar k$, so that not all automorphisms of $\bar k/k$ fix $\alpha$, and we just need to check we can restrict this automorphism to $L$. But one of the standard formulations of the condition that $L/k$ be normal is that every automorphism of $\bar k$ over $k$ map $L$ into itself.
