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 $F[x].$

Question

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.

Answer 1

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 $\sigma'(F) = F$.