Lemma 63.15.9. Let $P$ be a $\Lambda [\Gamma ]$-module, finite projective as $\Lambda [G]$-module. Then the coinvariants $P_ G = \Lambda \otimes _{\Lambda [G]} P$ form a finite projective $\Lambda $-module, endowed with an action of $\Gamma /G = \mathbf{N}$. Moreover, we have

\[ \text{Tr}_\Lambda (1; P_ G) = \sum \nolimits '_{\gamma \mapsto 1} \text{Tr}_\Lambda ^{Z_\gamma }(\gamma , P) \]

where $\sum _{\gamma \mapsto 1}'$ means taking the sum over the $G$-conjugacy classes in $\Gamma $.

**Sketch of proof.**
We first prove this after multiplying by $\# G$.

\[ \# G\cdot \text{Tr}_\Lambda (1; P_ G) = \text{Tr}_\Lambda (\sum \nolimits _{\gamma \mapsto 1} \gamma , P_ G) = \text{Tr}_\Lambda (\sum \nolimits _{\gamma \mapsto 1} \gamma , P) \]

where the second equality follows by considering the commutative triangle

\[ \xymatrix{ P^ G \ar[rd]_ a & & P_ G \ar[ll]^ c \\ & P \ar[ur]_ b } \]

where $a$ is the canonical inclusion, $b$ the canonical surjection and $c = \sum _{\gamma \mapsto 1} \gamma $. Then we have

\[ (\sum \nolimits _{\gamma \mapsto 1} \gamma ) |_ P = a \circ c \circ b \quad \text{and}\quad (\sum \nolimits _{\gamma \mapsto 1} \gamma ) |_{P_ G} = b \circ a \circ c \]

hence they have the same trace. We then have

\[ \# G\cdot \text{Tr}_\Lambda (1; P_ G) = {\sum _{\gamma \mapsto 1}}' \frac{\# G}{\# Z_\gamma }\text{Tr}_\Lambda (\gamma , P) = \# G{\sum _{\gamma \mapsto 1}}' \text{Tr}_\Lambda ^{Z_\gamma }(\gamma , P). \]

To finish the proof, reduce to case $\Lambda $ torsion-free by some universality argument. See [SGA4.5] for details.
$\square$

## Comments (2)

Comment #3622 by Owen B on

Comment #3725 by Johan on