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The proof of theorem in title has already been sketched in a question on MathStack (here); I tried to write the detailed proof, and I want to check it.

Let $\phi_1,\cdots,\phi_n$ be all the Brauer characters of a finite group $G$ (so they have domain $G_{p'}=\{g\in G: p\nmid o(g)\}$, and values in $\mathcal{O}\subseteq \mathbb{C}$, the ring of algebraic integers; $p$ is the characteristic of field $F$ over which the inequivalent irreducible $F$-representations of $G$ are defined with above Brauer characters.

Claim: Suppose $\phi_1,\cdots,\phi_n$ are independent over $\overline{\mathbb{Q}}$; then they are independent over $\mathbb{C}$.

Proof: (1) First if $|G_{p'}|=m$, then $m$ should be $\leq n$, otherwise $\phi_i$'s can-not be independent.

(2) Let $G_{p'}=\{g_1,g_2,\cdots,g_m\}$. For each $g_i$ we associate a row-vector $$[\phi_1(g_i),\phi_2(g_i),\cdots,\phi_n(g_i)]\in \mathbb{\overline{Q}}^n.$$ We get $m$ vectors $\{v_{g_1}, v_{g_2},\cdots,v_{g_m}\}$ in $\overline{\mathbb{Q}}^n$, and $m\leq n$.

(3) The $\overline{\mathbb{Q}}$-independence of $\phi_i$'s is equivalent to say that these $m$ rows in $\overline{\mathbb{Q}}^n$ are independent over $\overline{\mathbb{Q}}.$ Extend this set to a basis of $\overline{\mathbb{Q}}^n$: $$\{v_{g_1}, v_{g_2},\cdots,v_{g_m}, w_{m+1},\cdots, w_n\}.$$ (4) The matrix $P$ formed by these $n$ vectors as rows of a matrix $M_n(\overline{\mathbb{Q}})$ will be invertible.

(5) Hence the matrix $P$ as an element of $M_n(\mathbb{C})$ will be invertible.

(6) Hence $\phi_1,\cdots,\phi_n$ as functions into $\mathbb{C}$ are $\mathbb{C}$-independent.

Is this proof correct?

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Your proof looks fine to me. One thing I'll point out is that this really has nothing to do with the fact that you are dealing with Brauer characters.

You have some vectors $v_1, v_2, ... , v_n$ in a finite dimensional vector space $V$ (here class functions on $G_{p'}$) over one field $k$ (in this case $\bar{\mathbb{Q}}$) that are linear independent and you want to make sure they remain linearly independent when you extend scalars to a field extension $K$ ($\mathbb{C}$ in this case). More precisely you want to know that $v_1\otimes 1, v_2 \otimes 1, ... $ are linearly independent inside $V \otimes K$. Your proof works fine in this level of generality.

I'll note that in fact $V$ need not be finite dimensional for this result to hold, your proof would need to be modified a bit for the infinite dimensional case (as you complete to a basis and show a certain matrix is invertible) but it's not that hard to fix.

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