My questions is about this famous result:
Let $L\supset N \supset F$ be a tower of field extensions where $N$ is normal. For any $\sigma \in Gal(L/F), \sigma(N)=N$.
The following is what I have done which may not be helpful
It suffices to show that $\sigma(N)\subset N$. Let $\{u_i\}$ be a basis of $N/F$ and $f_i\in N[x]$ be the minimal polynomial of $u_i$ by the algebraicity of $N/F$. It follows from the normality of $N$ that $N$ is a splitting field of $\{f_i\}$. The above is the only thing I can get from the normality and I can't proceed from here.
This result proves it after restricting the homomorphisms in $Gal(L/F)$ to $N$ and seeing that $L \subseteq \overline{N}$:
Proof:
Let $\{u_i | i \in I \}$ be a generating system of $N/F$. Let $f_i = \mu_{x_i, F}$. Since $N/F$ is normal, those polynomials split over $N$. Let $R$ be the set of all roots of those polynomials. Then $F(R)=N$.
For any $x \in R$ and any $F$-homomorphism $\phi$, it holds that $\phi(x)$ will be a root of some $f_i$. Since $N=F(R)$ it follows that $\phi(N) \subseteq N$.
If $x_1, ..., x_n$ are the roots of some $f_i$, then any $F$-homomorphism $\phi$ will permute those roots, so $\phi(x_1), ..., \phi(x_n)$ are roots of $f_i$. Thus $N \subseteq \phi(N)$.
Let me know if everything is clear.