Lemma 64.15.3. Let f : P\to P be an endomorphism of the finite projective \Lambda [G]-module P. Then
\text{Tr}_{\Lambda }(f; P) = \# G \cdot \text{Tr}_\Lambda ^ G(f; P).
Proof. By additivity, reduce to the case P = \Lambda [G]. In that case, f is given by right multiplication by some element \sum \lambda _ g\cdot g of \Lambda [G]. In the basis (g)_{g \in G}, the matrix of f has coefficient \lambda _{g_2^{-1}g_1} in the (g_1, g_2) position. In particular, all diagonal coefficients are \lambda _ e, and there are \# G such coefficients. \square
Comments (0)