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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)