Q:

Why must a symmetric polynomial of the conjugates lie in $\mathbb{Q}$? 

Let $K/\mathbb{Q}$ be an algebraic number field of degree $n$, and suppose $\alpha \in K.$ Let $\sigma_i \colon K \to \mathbb{C}$ for $i = 1,\dots,n$ be the $n$ embeddings.  Why is it that symmetric polynomials of the conjugates $\sigma_i(\alpha)$ must be in $\mathbb{Q}?$ Is it because if $m(x)\in \mathbb{Q}[x]$ is the minimal polynomial of $\alpha$, then its coefficients, which are the elementary symmetric polynomials $s_i$ of the conjugates, must be in $\mathbb{Q}$, hence by the fundamental theorem of symmetric polynomials, any symmetric function of $\sigma_i(\alpha)$ is in $\mathbb{Q}[s_1,\dots,s_n]$? 

A:

The symmetric polynomials of the conjugates $\sigma_i(\alpha)$ are not the coefficients of the minimal polynomial of $\alpha$, but they are the coefficients of the characteristic polynomial of $\alpha$ w.r.t the extension $K/\mathbb Q$ (which is the characteristic polynomial of the $\mathbb Q$-linear map $K\to K$ given by multiplication with $\alpha$).  In fact this happens to be a power of the minimal polynomial (and the exponent is $1$ if $\alpha$ generates $K$). 
With that said, these are for sure rational numbers.