Field Extensions Problem - From Paolo Aluffi's Book This is the exercise 4.7, chapter VII, from Paolo Aluffi's algebra book.
I'm sorry for just copying the question without writing any development myself, I don't have a single ideia about how to use the definitions or the theorems given to solve this. 
I need the answer but more than that I need to understand the answer, slow explanations are very welcome ( I'm slow at this subject ).

Let $k\subseteq F=k(\alpha)$ be a simple algebraic extension. Prove that $F$ is normal over $k$ if and only if for every algebraic extension $F\subseteq K$ and every $\sigma\in \rm{Aut \ }_k(K)$, $\sigma(F)=F$.

$\rm{Aut \ }_k(K)$ is the group of automorphisms of $K$ fixing the field $k$.
Thanks in advance.
 A: Assume that $F = k(\alpha)/k$ is normal. Let $f \in k[X]$ be the minimal polynomial of $\alpha$ and let $\alpha_1,\ldots,\alpha_n$ be its roots, i.e. the conjugates of $\alpha$. By assumption  $\alpha_1,\ldots,\alpha_n \in k(\alpha)$. Let $K/k(\alpha)$ be an algebraic extension and let $\sigma \in \operatorname{Aut}_k(K)$. Since $\sigma$ fixes $k$, we have $$f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0,$$
i.e. $\sigma(\alpha) = \alpha_i$ for some $i$. So $\sigma(\alpha) \in F$ and hence $\sigma(F) \subseteq F$ for all $\sigma \in \operatorname{Aut}_k(K)$. Since $\sigma^{-1}(F) \subseteq F$, it follows that $F \subseteq \sigma(F)$, hence $\sigma(F) = F$ for all $\sigma \in \operatorname{Aut}_k(K)$.
For the converse, we need the following fact:

Theorem: Let $M/L/k$ be algebraic extensions and let $K$ be an algebraically closed field containing $k$. Then any $k$-embedding $L \hookrightarrow K$ can be extended to an embedding $M \hookrightarrow K$.

Now assume that $\sigma(F) = F$ for every $\sigma \in \operatorname{Aut}_k(K)$ where $K/k(\alpha)$ is any algebraic extension.
Let $K$ be an algebraic closure of $k(\alpha)$ and let $\alpha_1,\ldots,\alpha_n \in K$ be the conjugates of $\alpha$. The fields $k(\alpha_i)$ are all isomorphic to $k(\alpha)$ over $k$ (both are isomorphic to $k[X]/(f)$ where $f$ is the minimal polynomial of $\alpha$). By the quoted theorem, the composite map $k(\alpha) \to k(\alpha_i) \hookrightarrow K$ can be extended to a $k$-homomorphism $\sigma: K \to K$. Since $K/k$ is algebraic, it is an automorphism. We have $\sigma(\alpha) = \alpha_i$, so by assumption $\alpha_i \in F$. So $F$ contains all conjugates of $\alpha$, i.e. $F/k$ is normal.
