Let $K$ be a global field, $f(x)\in K[x]$ be an irreducible separable polynomial and $L$ be the splitting field of $f(x)$. Suppose that the Galois group of $L$ over $K$ is the symmetric group $S_{\deg(f)}$. Do you have a simple way to find a defining equation for $L$ in terms of the coefficients of $f$? Essentially what I need is a closed expression for the minimal polynomial of a primitive element (in terms of the coefficients of $f(X)$). I believe this should involve some symmetric polynomial things...
In principle, you get $L$ simply by progressively extending $K$ with the roots of $f(x)$: you will get a sequence $K=L_0\subseteq L_1 \subseteq L_2 \subseteq \dots \subseteq L_{\deg(f)}=L$, where $[L_{i+1}:L_i]=\deg(f)-i$ (this follows because the Galois group is $S_n$, so the roots are essentially indistinguishable). In some sense we know the equation at each step just by euclidean division. Now, I am not sure how to proceed in order to avoid terrible computations, maybe somebody has a nice idea.
If you find it more convenient, you might replace $K$ with $k(t)$, where $k$ is a finite field.