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My questions is the following.

Let $G$ be a finite group and let $M$ be a $\mathbb{Q}G$ module. If $$\mathcal{N}\subset \mathbb{C}\otimes_\mathbb{Q} M$$ is a $\mathbb{C}G$-submodule of $\mathbb{C}\otimes_\mathbb{Q} M$, does there exist a finite field extension $k$ of $\mathbb{Q}$ and a $kG$-submodule $$N_k\subset k\otimes_\mathbb{Q}M$$ such that $\mathbb{C}\otimes_k N_k = \mathcal{N}$?

I'm hoping that such a $N_k$ exists and that we can take $k$ to be $\mathbb{Q}$ adjoin all the $|G|$th roots of unity. I cannot think of a proof of this or come up with a counter example.

My hope comes from the fact in linear algebra that if $A$ is an $n\times n$ matrix with entries in $\mathbb{Q}$ and $\lambda$ is an eigenvalue of $A$, then $\mathbb{Q}(\lambda)$ is a finite extension of $\mathbb{Q}$ and an eigenvector of $A$ with eigenvalue $\lambda$ is defined over $\mathbb{Q}(\lambda)$. Furthermore, for a finite group $G$, and a $G$-module $M$, all the eigenvalues of elements of $G$ acting on $M$ are $|G|$th roots of unity.

Any help with this, or a reference that explains how to answer the problem would be greatly appreciated!!

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I figured out the answer to my question (maybe I should just delete the question, but I'll put the answer here). The answer is no by the following example. Let $G$ be the cyclic group of order 2. Then $G$ acts on $\mathbb{Q}^2$ by letting the generator of $G$ act by scalar multiplication by -1. Let $\mathcal{N}\subset\mathbb{C}^2$ be the one dimensional subspace spanned by $(1, \pi)$. Since $\pi$ is transcendental, there does not exist a finite extension $k$ of $\mathbb{Q}$ such and a one dimensional subspace $N_k\subset k^2$ such that $\mathbb{C}\otimes N_k = \mathcal{N}$.

I believe it is true that if $\mathbb{C}\otimes M$ decomposes as the direct sum of distinct irreducible submodules (so no irreducible representation of $G$ shows up more than once in the decomposition of $\mathbb{C}\otimes M$), then the answer to the question is yes.