Nagao error terms have zero trace on the relevant p-section

The subgroup $D=⟨u⟩$ is central in $H$, and $D{C}_G(D)=C_G(u)=H\le N_G(D)$, so F2 applies. Decompose the finite-rank lattice $M_{\mathrm{e}\mathrm{r}\mathrm{r}}$ into indecomposable $\mathcal{O}H$-lattices $U_1,\dots ,U_r$. If $u=1$, then $D=1$ and F2 says no vertex of an error summand contains $1$, which is impossible; thus $r=0$ and the result is immediate.

Suppose $u\mathrm{\ne }1$. For each $U_i$, choose a vertex $Q_i$. By F2, $D\le ̸Q_i$. Since $u$ is central in $H$, every $H$-conjugate of $u$ is $u$ itself; hence $u$ is not $H$-conjugate to an element of $Q_i$. Because $u$ and the $p$-regular element $v$ commute, F1 says that the $p$-part of $uv$ is exactly $u$. Each $U_i$ is relatively $Q_i$-projective by the definition of a vertex, so F3 gives $\mathrm{tr}(uv\mid K{\otimes }_{\mathcal{O}}{U}_i)=0.$

Scalar extension preserves the finite direct sum, and trace is additive. Step 1.2 proves the more precise assertion in the Statement. Therefore the character of $M_{\mathrm{e}\mathrm{r}\mathrm{r}}$ is zero at $uv$. Taking traces in $M=M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}}\oplus M_{\mathrm{e}\mathrm{r}\mathrm{r}}$ gives the required equality. The case $M_{\mathrm{e}\mathrm{r}\mathrm{r}}=0$ is the empty finite sum, already covered, and algebraic closedness is used exactly through F3.