Irreducible Brauer characters are independent on p-regular elements

Consider any finite relation ${\sum }_{i=1}^r{c}_i{\phi }_i=0$ over $K$. If some coefficient is nonzero, let $b={\min }_{c_i\mathrm{\ne }0}v(c_i)$ and replace every $c_i$ by ${\pi }^{-b}{c}_i$. Then all coefficients lie in $\mathcal{O}$ and at least one has valuation zero, hence has nonzero residue.

Reduce the relation at every p-regular element. It becomes ${\sum }_i{\bar{c}}_i\mathrm{tr}(s\mathrm{\mid }{S}_i)=0$. For arbitrary $g$, replace each trace by its trace at the same p-regular part $s$ of $g$. Thus ${\sum }_i{\bar{c}}_i\mathrm{tr}(g\mathrm{\mid }{S}_i)=0$ for every $g\in G$.

Since k splits G, modular trace independence forces every ${\bar{c}}_i=0$, contradicting step 1.1. Hence the original relation has all coefficients zero. This includes the empty family.

For any finite family the matrix of values on the finite p-regular set has independent columns. Successive elimination using nonzero pivots therefore supplies a square minor of full column size with nonzero determinant. That determinant remains nonzero under any field embedding, proving independence after extension. All entries lie in the subfield $E\subseteq K$ generated over $\mathbb{Q}$ by the finitely many lifted roots used here; these are roots of unity, so E is cyclotomic. The same minor lies in E and remains nonzero under a fixed embedding $E\to \mathbb{C}$. No embedding of all of K into C is assumed.