I'm re-reading the fundamental theorem of field extensions. (K is normal $\iff$ K is a factorization field.)
Assume $K=F(\alpha_1, \dots , \alpha_n)$, is the factorization field of $f\in F[x]$, over the field $F$. Then for each $\alpha\in K, \alpha = g(\alpha_1,\dots,\alpha_n)$, for some $g\in F[x_1\dots,x_n]$. Finally define $h\in K[x]$ by $$h=\prod_{\sigma\in S_n}(x-g(\alpha_{\sigma(1)},\dots,\alpha_{\sigma(n)}))$$ Here $S_n$ is the symmetric group of order $n!$ .
It was particularly hard for me to reason through proving that the coefficients of $h$ are polynomials (with coefficients in $F$) of the elementary symmetric polynomials of $\alpha_1, \dots , \alpha_n$. Furthermore by applying Vietas formulas for $f$ we find that its coefficients are symmetric polynomials (with integer coefficients) of its roots: $\alpha_1, \dots , \alpha_n$. My book hereby concludes that $h\in F[x]$.
I may be just tired by now but I feel like there is a step or two missing in the above conclusion. Any comments would be appreciated.
Notice that the Vieta formula for each coefficients is unchanged when we permute the $\alpha_i$ by any element of the symmetric group. This means that each coefficient is a symmetric polynomial.
But all symmetric polynomials of the roots of a polynomial $f$ have their values in the base field. This is essentially the converse of Vieta's formulas.