Lifted modular trace on p-regular elements

The polynomial $X^m-1$ splits in $k$. Indeed, for any monic irreducible factor $h$, the field $k[X]\mathrm{/}(h)$ is a simple module for $k⟨g⟩$; multiplication by its elements gives module endomorphisms. The scalar-endomorphism condition forces this field to equal $k$, so $\mathrm{deg}h=1$. Since the derivative $m{X}^{m-1}$ has no common root with $X^m-1$, the roots are distinct.

For each root $\lambda$ put $P_{\lambda }(X)={\prod }_{\nu \mathrm{\ne }\lambda }(X-\nu )\mathrm{/}(\lambda -\nu )$. Polynomial interpolation gives ${\sum }_{\lambda }{P}_{\lambda }(X)=1$ and $(X-\lambda )P_{\lambda }(X)=0$ modulo $X^m-1$. Applying these identities to $\rho (g)$ expresses $V$ as the direct sum of its eigenspaces: the sum spans, and application of $P_{\lambda }(\rho (g))$ isolates each summand. Thus all displayed eigenvalues lie in $k$ and have mth power one.

The unique lifts exist in $\mathcal{O}$, and their multiset depends only on the characteristic polynomial. A change of basis or conjugation of $g$ conjugates its matrix and preserves that polynomial, hence preserves the sum. For $V=0$ the sum is zero; at $g=1$ it is $d\cdot 1_{\mathcal{O}}$. These prove the stated well-definedness.