my question can be seen as a extension to this question.
Let $\overline{\mathbb{Q}}$ denote an algebraic closure of $\mathbb{Q}$. Given a polynomial with algebraic coefficients $f \in \overline{\mathbb{Q}}[X_1,...,X_n]$, say $$f=\sum_{(\alpha_1,...,\alpha_n)}c_{(\alpha_1,...,\alpha_n)}X^{\alpha_1}\cdot...\cdot X^{\alpha_n}.$$ Let $(b_1,...,b_n)$ be a root of this polynomial, where $b_1,...b_n \in \mathbb{C}$ (so the $b_{i}$ are not necessarily algebraic numbers individually). By root i mean that we have $$f(b_1,...,b_n) = \sum_{(\alpha_1,...,\alpha_n)}c_{(\alpha_1,...,\alpha_n)}b^{\alpha_1}\cdot...\cdot b^{\alpha_n} = 0.$$ Is it possible to construct a polynomial with rational coefficients $g \in {\mathbb{Q}}[X_1,...,X_n]$, for which $(b_1,...,b_n)$ is a root?
I would already be happy with hints to relevant literature. Thank you!
Here is a try. Let $L/Q$ be a finite Galois extension that contains all the coefficients of $f$. Denote by $G$ the corresponding Galois group. Take $$ g={\rm Norm}_{L/Q}(f) = \prod_{\sigma \in G} \sigma(f), $$ where by $\sigma(f)$, I mean the polynomial one gets by applying $\sigma$ to the coefficients of $f$. Then the coefficients of $g$ lie in $Q$, since they are invariant to the Galois action and clearly $$ g(b_1,\ldots,b_n) = f(b_1,\ldots, b_n) \prod_{1\neq \sigma\in G} \sigma(f)(b_1,\ldots, b_n) = 0 $$ as needed.
Edit: More details. Write $f = \sum c_{\alpha_1, \ldots, \alpha_n} X^{\alpha_1}\cdots X^{\alpha_n}$. Then $\sigma(f) = \sum \sigma( c_{\alpha_1, \ldots, \alpha_n}) X^{\alpha_1}\cdots X^{\alpha_n}$. Now, write $$ g = \prod_{\sigma \in G} \sigma(f) = \sum d_{\alpha_1,\ldots, \alpha_n} X^{\alpha_1}\cdots X^{\alpha_n}. $$ Note that, on the one hand $\tau (g) = \prod_{\sigma\in G} \tau\sigma(f) = \prod_{\rho\in G} \rho(f) = g$ (I've substituted $\rho = \tau\sigma$ and noted that as $\sigma$ runs over the elements of $G$, so does $\rho$). On the other hand, $\tau(g) = \sum \tau(d_{\alpha_1,\ldots, \alpha_n}) X^{\alpha_1}\cdots X^{\alpha_n}$. Therefore, by comparing coefficients, we get that $\tau(d_{\alpha_1,\ldots, \alpha_n}) = d_{\alpha_1,\ldots, \alpha_n}$ for all $\tau\in G$ and $\alpha_1,\ldots, \alpha_n$, so $d_{\alpha_1,\ldots, \alpha_n}\in Q$.