I was told it is possible to prove splitting fields are normal using the Fundamental Theorem of Symmetric Polynomials rather than the usual approach. Does anyone have hints or a reference for this?
something like:
Let $\alpha_i$ be roots of a polynomial and $L = K(\alpha_1,…,\alpha_n)$. If $m(x) \in K[x]$ is an irreducible polynomial with a root $\beta$ in $L$, then there is some polynomial $g(x_1, x_2, \ldots, x_n) \in K[x_1, x_2, \ldots, x_n]$ such that $\beta = g(\alpha_1, \alpha_2, \ldots, \alpha_n)$. Then I need to show $m$ splits. I think it has something to do with letting $$(\sigma g)(x_1, x_2, \ldots, x_n) := g(x_{\sigma(1)}, x_{\sigma(2)}, \ldots, x_{\sigma(n)}$$ for $\sigma \in S_n$ but I'm unclear where to go
Best Answer
If $L/{K}$ is not separable then the $\alpha_{j}$ must be repeated according to their multiplicity in the polynomial $h\in K[x]$ whose $L$ is the splitting field.
Consider $$f=\prod_{\sigma\in S_n} (x-\sigma g(\alpha_1,\ldots,\alpha_n))=\sum_{m=1}^{n!} c_m x^m$$ Its coefficients $c_m$ are some polynomials in the $\alpha_j$ that are invariant under the action of $S_n$. So they lie in the field extension of $K$ generated by the coefficients of $h=\prod_j (x-\alpha_j)$, ie. $c_m\in K, f\in K[x]$. Here we can take $g(\alpha_1, \ldots, \alpha_n) = k_1 \alpha_1 + \ldots + k_n \alpha_n$ for some $k_i \in K$.
As $m$ divides $f$ (since $m$ is a minimal, hence irreducible, polynomial over $K$ and one of its roots $\beta = g(\alpha_1, \ldots, \alpha_n)$ is a root of $f$), and $f$ splits completely in $L$ it means that $m$ splits completely in $L$.