# Prove that $F$ is normal over $K$ iff for any irreducible $f(x)\in K[x],$ if $f$ has a root in $F$ then $f$ splits in $F[x].$

Let $K\subseteq F$ be a tower of fields. We say that $F$ is normal over $K$ if $F$ is a splitting field of some set of polynomials in $K[x].$

I need to prove that if the above tower is algebraic then $F$ is normal over $K$ if and only if for any irreducible $f(x)\in K[x],$ if $f$ has a root in $F$ then $f$ splits in $F[x].$

I could prove the reverse implication by considering the set $S=\{\min(K,a): a\in F\}$, where $\min(K,a)$ is the minimal polynomial of $a$ over $K$. However I'm having trouble trying to think of a start to prove the forward implication. The assumption $F$ is normal over $K$ grants only a set of polynomials over $K$ for which $F$ is a splitting field. How do I connect "for any irreducible $f(x)\in K[x],$ if $f$ has a root in $F$ then $f$ splits in $F[x]$"? Could someone please give me a hint? Thanks.

Suppose $$F$$ is the splitting field for some set of polynomials in $$K[x]$$. Let $$X$$ be the set of the roots of these polynomials. Thus, we have $$K(X) = F$$. Any embedding of $$F$$ into $$\bar{K}$$ which fixes $$K$$, namely, $$\sigma: F \to \bar{K}$$ permutes the elements of $$X$$ (let me know if you didn't know this). Since $$\sigma(X) = X$$ and $$\sigma(K) = K$$, we have that $$\sigma(F) = F$$.
Now suppose towards a contradiction that there exists an irreducible polynomial in $$K[x]$$ with roots $$\alpha , \beta$$ and $$\alpha \in F$$ and $$\beta \notin F$$. We know there exists an embedding $$\sigma: K(\alpha) \to \bar{K}$$ which fixes $$K$$ and where $$\alpha \to \beta$$. Since $$F$$ is algebraic over $$K(\alpha)$$, this extends to an embedding $$\sigma ': F \to \bar{K}$$ such that $$\sigma'_{|K(\alpha)} = \sigma$$ and hence fixes $$K$$.
But then $$\sigma'(\alpha) = \beta$$ and $$\beta \notin F$$ so this contradicts that $$\sigma'(F) = F$$.