I'm trying to prove that if $K$ is a finite field extension of $F$ such that $K$ is the splitting field of some collection $C$ of polynomials in $F[x]$, then every irreducible polynomial in $F[x]$ with a root in $k$ splits completely in $K$.
So let $f\in K[x]$ be an irreducible polynomial with roots $\alpha_1,\ldots,\alpha_n$ in an algebraic closure, with $a_1\in K$. I was to show that the other $\alpha_i$'s are also in $K$.
Since $\alpha_1\in K$, it can be written as a rational function over $F$ using a finite number of the roots of the polynomials of $C$. So say $\alpha_1=g(\beta_1,\ldots,\beta_m)$. Since $\alpha_i$ is another root of the irreducible polynomial $f$, there is an automorphism of the algebraic closure that sends $a_1\mapsto a_i$. This automorphism $\sigma$ must also send roots of polynomials in $C$ to roots of the same polynomials, which means that $a_i=g(\beta_1',\ldots,\beta_m')$, where $\beta_i'=\sigma(\beta_i)$. But since the $\beta_i'$'s are in the splitting field $K$, so is $a_i$, as desired.
Is my proof correct?