Rationality of field embeddings

Question

After my earlier question question turned out to have a negative answer (Thank you to all respondents!), here is a more modest one. Both a positive answer and a counterexample would help my work. If some context is of interest, these questions turned up while studying self-similar measures with algebraic contraction ratios. The sets $S$ I'm interested in are of the form $S_\lambda=\{\sigma: |\sigma(\lambda)|<1\}$ for some $\lambda$ generating $K$, if it makes any difference.

Question: Let $K$ be a finite extension of $\mathbb{Q}$. Let $S$ be a subset of the field embeddings into $\mathbb{R}, \mathbb{C}$ (picking both or none in a pair of complex embeddings). If $$\sum_{\sigma \in S} \sigma(x) \in \mathbb{Q}$$ for all $x \in K$, does $S$ then necessarily consist of all (or no) field embeddings?

Answer 1

The answer is yes. Suppose $S$ is nonempty. Write $K=\mathbb Q(\alpha)$ (using the primitive element theorem). Applying your assumption to $\alpha^n$ for all $n\in\mathbb N$ we get that all sums $\sum_{\sigma\in S}(\sigma(\alpha))^n$ are rational. Using Girard-Newton formulas, this implies that all the elementary symmetric polynomials in $\{\sigma(\alpha)\mid\sigma\in S\}$ are rational. This implies that the polynomial $\prod_{\sigma\in S}(x-\sigma(\alpha))$ has rational coefficients. However, each $\sigma(\alpha)$ has degree $[K:\mathbb Q]$ over $\mathbb Q$, so the degree $|S|$ of this polynomial must be at least $[K:\mathbb Q]$, which means precisely that $S$ consists of all embeddings $\sigma$.