# Brauer characters: independence over $\overline{\mathbb{Q}}$ implies independence over $\mathbb{C}$

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?

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.