Prove that if $F$ is a splitting Field of $S$ over $K$ and $E$ is an intermediate field then $F$ is a splitting field of $S$ over $E$.

Question

This happens to be Hungerford problem 5.3.2. Here $S$ is a set of polynomials in $K[x]$. $F$ is a splitting field of $S$ over $E$ if

1. Every $f\in S$ splits in $F$
2. $F= E(X)$ where $X=\{ \text{roots of all polynomials in }$ S}

Property 1 follows quickly from the fact that $K[x] \subset E[x]$ but I am stuck on property 2. By assumption, we know that $F=K(X)\subseteq E(X)$ so $F\subseteq E(X)$. For the reverse inclusion I am stuck and in general I don't see why it should be true.

Answer 1

$E \subseteq F$ because it's an intermediate field and $X \subseteq F$ because $X$ all the roots of polynomials in $S$, since $X \subseteq K(X) = F$, by hypothesis.

Therefore $F= K(X) \subseteq E(X) \subseteq F$, since $E$ and $X$ both are contained in F therefore the field generated by them.