Let $w$ be a primitive $p^{th}$ root of unity, where $p$ is a prime and $n \geq 3$ is a positive integer. I am trying to determine if there exists integers $\alpha_{1}, \alpha_{2}, \ldots , \alpha_{n}$ such that $$ \big( \alpha_{1} \sum_{i \in J_{1}} w^{i} + \alpha_{2} \sum_{i \in J_{2}} w^{i} + \ldots + \alpha_{n} \sum_{i \in J_{n}} w^{i} \big) \not\in\mathbb{Q} $$ for every $J_{1}, J_{2}, \ldots, J_{n} \subset \{1, 2, \ldots, p-1\}$ such that at least one of the $J_{i}$ is not equal to $\{1, 2, \ldots, p-1\}$.
My attempt
Assume for the contradiction that there is a relation of the form $$ \alpha_{1} \sum_{i \in J_{1}} w^{i} + \alpha_{2} \sum_{i \in J_{2}} w^{i} + \ldots + \alpha_{n} \sum_{i \in J_{n}} w^{i} = q \hspace{3mm} (*)$$ for some rational number $q$ and some non-zero integers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} \in\mathbb{Z}\setminus\{0\}$. We can rewrite this relation by collecting coefficients associated with powers $w, w^{2}, \ldots, w^{p-1}$ as
$$ w^{1} \sum_{i \in I_{1}} \alpha_{i} + w^{2} \sum_{i \in I_{2}} \alpha_{i} + \ldots + w^{p-1} \sum_{i \in I_{p-1}} \alpha_{i} = q \hspace{3mm} (*) $$ for subsets $I_{1}, I_{2}, \ldots, I_{p-1} \subset \{1, 2, \ldots, n\}$ such that at least one of the $I_{\nu}$ is not equal to $\{1, 2, \ldots, n\}$. Since the minimal polynomial of $w$ is $1 + w + \ldots + w^{p-1}$, the relation $(*)$ implies that
$$ \sum_{i\in I_{1}} \alpha_{i} = \sum_{i\in I_{2}} \alpha_{i} = \ldots = \sum_{ i \in I_{p-1}} \alpha_{i}. \hspace{3mm} (**) $$ We want to show that $(**)$ is not satisfied for some $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})\in\mathbb{Z}^{n}$. One way $(**)$ can hold is when $I_{1} = I_{2} = \ldots = I_{p-1} = I (say) \subset\{1, 2, \ldots, n\}$. Since $\alpha_{i}$ were non-zero, i.e., each $\alpha_{i}$ appears in some $I_{j}$, this implies $I = \{1, 2, \ldots, n\}$ and hence $J_{1} = J_{2} = \ldots = J_{n} = \{1,2,\ldots, p-1\}$, contradicting our initial assumption.
On the other hand, if $I_{r} \neq I_{s}$ for some $r, s\in\{1, 2, \ldots, p-1\}$, then consider the equality $$\sum_{i \in I_{r}} \alpha_{i} = \sum_{i \in I_{s}} \alpha_{i}$$ which gives rise to an equation of the form $$\sum_{m=1}^{n} \epsilon_{m} \alpha_{m}=0,$$ where $\epsilon_{m}$ are in $\{-1,0,1\}$; the solution set of which has dimension at most $(n-1)$ in $\mathbb{Z}^{n}$. Therefore, there are $(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n})$ not satisfying $(**)$ and hence not satisfying $(*)$.
For every such $(J_1,\ldots,J_n)$, let $V_{J}$ be the set of $\alpha \in \mathbb{Q}^n$ such that $\sum_{i=1}^n{\alpha_i\sum_{t \in J_i}{w^t}}\in \mathbb{Q}$. Then $V_{J}$ is a proper sev of $\mathbb{Q}^n$.
A large multiple of some $\alpha$ not in the reunion of the $V_J$ (there are finitely many of them so their reunion is not all of $\mathbb{Q}^n$) then works.