Let $n\in\mathbb{N}$. Is there a way to study the set $\mathcal{S}_{n}$ of $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})\in\mathbb{C}^{n}$ (resp. $\in\mathbb{R}^{n}$), such that the polynomial $$(x-\alpha_{1})(x-\alpha_{2}) \cdots (x-\alpha_{n})$$ is in $\mathbb{Q}[x]$? Of course, it is same as studying the set of $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})\in\mathbb{C}^{n}$ (resp. $\mathbb{R}^{n}$) such that $e_{i}(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}) \in\mathbb{Q}$ for every $1 \leq i \leq n$. Here $e_{i}$ is the $i^{th}$ elementary symmetric polynomial in $n$ variables. $\mathbb{Q}^{n}$ is already in this set, but what else?
I am not even sure what methods, if any, one would use to study such a question. Do you know of any results number-theoretic, combinatorial, geometric or arithmetic about the set $\mathcal{S}_{n}$ in the literature?
I am not an expert on the topic but here are some observations:
First of all since all the $\alpha_i$ are roots of rational polynomials we have $\mathbb{Q}^n \subset \mathcal{S}_n \subset \mathbb{A}^n$ where $\mathbb{A}$ denotes the algebraic integers. If you have $\alpha \in \mathcal{S}_n$ then for any $\alpha_i$ the minimal polynomial $f_{\alpha_i}$ over $\mathbb{Q}$ divides $(x-\alpha_1) \cdots (x-\alpha_n)$, so any conjugate of $\alpha_i$ must be among the $\alpha_j$ exactly once (or $k$ times if you have $k$ copies of $\alpha_i$). And as far as I can tell this is all the structure your set has. You can generate a member of $\mathcal{S}_n$ by taking any rational polynomial $f_1$ with degree $d_1 \leq n$ and add all it's roots to your list. Then you pick another rational polynomial with degree $d_2 \leq n-d_1$ and add all it's roots to your list. You keep doing this until $d_1 + \cdots +d_k = n$ so that you have $n$ roots collected in total. Any member of your set can be generated like this.
To be honest I don't see a point in using a geometric or combinatorial approach if standard Galois theory describes the set just fine. Or perhabs you need to clarify what exactly you mean by
studying the set.