# If $E$ is the splitting field of a separable polynomial on $K$, then $E/K$ is normale.

Let $K$ a field and $E$ the splitting field of a separable polynomial $f\in K[X]$. Show that $E/F$ is normale.

My definition of normale extension is that $E/K$ is normale if $$Hom_K(E,K^{alg})=Aut_K(E).$$

This is how I do. Let $f\in K[X]$ a separable polynomial that split over $E$ and let $\alpha \in E$ a root of $f$. Since $E$ is the splitting field of $f$, $E$ is generated by the roots of $f$. Let denote $\alpha _1,...,\alpha _n$ the roots of $f$. Since $\sigma$ permute the roots of $f$, and that every element of $E$ is a combinaison of $\alpha _1,...,\alpha _n$, we clearly have that $\sigma (x)\in E$ for all $x\in E$.

Is this proof correct ? The end is not very rigorous, do you have a better idea ?

You have to show that every $s:K\rightarrow K^{alg}$ $s(K)=K$. but since $f^s=f$, $s$ preserves the set of roots of $K$, so $s(K)$ contains $K$ since by definition $K$ is the small field where $f$ splits; this implies that $s(K)=K$ since they have the same degree.