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

Question

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 $\mathbb{Q}[s_1,\dots,s_n]$?

Answer 1

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.

Answer 2

I totally agree with MooS. Here is an alternative argument:

Since $\alpha \in \mathbb Q(\theta)$, it is clear that $\alpha = p(\theta)$ for some polynomial $p(X) \in \mathbb Q[X]$. But then $\sigma_i(\alpha) = p(\sigma_i(\theta))$ for each monomorphism $\sigma_i$. So every symmetric polynomial of the $\sigma_i(\alpha)$'s is also a symmetric polynomial of the $\sigma_i(\theta)$'s! The $\sigma_i(\theta)$'s are distinct, and they are precisely the roots of the minimal polynomial of $\theta$. So by the fundamental theorem of symmetric polynomials, every symmetric polynomial of the $\sigma_i(\theta)$'s (and hence, every symmetric polynomial of the $\sigma_i(\alpha)$'s) is a polynomial of the coefficients of the minimal polynomial of $\theta$, and hence, is in $\mathbb Q$.

[As MooS pointed out, it is true that the $\sigma_i(\alpha)$'s are roots of the minimal polynomial of $\alpha$, but they are not necessarily distinct: the list $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$ may contain repetitions. (For example, if $\theta = \sqrt{2}$ and $\alpha = 5$, then $\sigma_1(\theta) = \sqrt{2}$ and $\sigma_2(\theta) = - \sqrt{2}$, but $\sigma_1(\alpha) = \sigma_2(\alpha) = 5$.) So if we want to argue directly with the $\sigma_i(\alpha)$'s, we need to do the extra work of showing that each root of the minimal polynomial of $\alpha$ appears the same number of times in the list $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$.]