Ordinary traces on p-singular elements are not determined by reduction

Take the rank-one lattices $L_{+}=\mathcal{O}$ and $L_{-}=\mathcal{O}$ with t acting by 1 and -1. These are representations since both scalars square to 1. Their reductions are both the trivial k-module because $-1=1$ in characteristic two. Their K-character values at t are 1 and -1, different because K has characteristic zero. The element t has order 2 and is p-singular, so this does not conflict with the domain of the Brauer character.

On $k^2$ take $J=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. Then $J^2=I$ in characteristic two, $J\mathrm{\ne }I$, and $\mathrm{tr}(J)=0=\mathrm{tr}(I)$. The p-regular part of t is 1. Thus the trace equality holds exactly as stated in the trace lemma while equality of operators fails. Both proposed conclusions have explicit counterexamples.