Symmetric Polynomials and Splitting Fields

The more symmetric $\alpha$ is, the closer it is to the base field $F$.

Symmetric Functions Theorem

Every symmetric polynomial $f(x_1,x_2,\dots ,x_n)\in R[x_1,\dots x_n]$ with coefficients in a ring $R$ can be written in a unique way as a polynomial in the elementary symmetric polynomials.

e.g.

Corollary

Suppose that a polynomial $f(x)=x^n+a_1{x}^{n-1}+\cdots +a_n$ has coefficients in a field $F$, and it splits completely in an extension field $K$, with roots ${\alpha }_1,\dots ,{\alpha }_n$. Let $g(u_1,\dots ,u_n)$ be a symmetric polynomial in $u_1,\dots ,u_n$ with coefficients in $F$. Then $g({\alpha }_1,\dots ,{\alpha }_n)\in F$.

Proof Symmetric functions theorem tells that $g$ is a polynomial in the elementary symmetric functions. Say that $g(u_1,\dots ,u_n)=\tilde{g}(s_1(u),\dots ,s_n(u))$, where $\tilde{g}\in F[x]$ as well. When we evaluate at $u=\alpha$, we obtain $s_i(\alpha )=\pm a_i$, so $g({\alpha }_1,\dots ,{\alpha }_n)=\tilde{g}(s_1(u),\dots ,s_n(u))$ must also be in $F$. $\square$

Splitting Fields

Splitting Theorem

Let $K\supset F$ be a splitting field extension of a polynomial $f\in F[x]$. If an irreducible polynomial $g\in F[x]$ has one root in $K$, then it splits completely in $K$.

The Discriminant