Proof.
Since $\mathbf{Z}/\ell ^ n\mathbf{Z}$ is a local ring and $K_ n$ is perfect, each $K_ n$ can be represented by a finite complex $K_ n^\bullet $ of finite free $\mathbf{Z}/\ell ^ n \mathbf{Z}$-modules such that the map $K_ n^ p \to K_ n^{p+1}$ has image contained in $\ell K_ n^{p+1}$. It is a fact that such a complex is unique up to isomorphism. Moreover $\pi _ n$ can be represented by a morphism of complexes $\pi _ n^\bullet : K_ n^\bullet \to K_ n^\bullet $ (which is unique up to homotopy). By the same token the isomorphism $\varphi _ n : K_{n+1} \otimes _{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}}^{\mathbf{L}} \mathbf{Z}/\ell ^ n\mathbf{Z}\to K_ n$ is represented by a map of complexes
\[ \varphi _ n^\bullet : K_{n+1}^\bullet \otimes _{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}} \mathbf{Z}/\ell ^ n\mathbf{Z} \to K_ n^\bullet . \]
In fact, $\varphi _ n^\bullet $ is an isomorphism of complexes, thus we see that
there exist $a, b\in \mathbf{Z}$ independent of $n$ such that $K_ n^ i = 0$ for all $i\notin [a, b]$, and
the rank of $K_ n^ i$ is independent of $n$.
Therefore, the module $K_\infty ^ i = \mathop{\mathrm{lim}}\nolimits _ n \{ K_ n^ i, \varphi _ n^ i\} $ is a finite free $\mathbf{Z}_\ell $-module and $K_\infty ^\bullet $ is a finite complex of finite free $\mathbf{Z}_\ell $-modules. By induction on the number of nonzero terms, one can prove that $H^ i\left(K_\infty ^\bullet \right) = \mathop{\mathrm{lim}}\nolimits _ n H^ i\left(K_ n^\bullet \right)$ (this is not true for unbounded complexes). We conclude that $H_\infty ^ i = H^ i\left(K_\infty ^\bullet \right)$ is a finite $\mathbf{Z}_\ell $-module. This proves ii. To prove the remainder of the lemma, we need to overcome the possible noncommutativity of the diagrams
\[ \xymatrix{ {K_{n+1}^\bullet } \ar[d]_{\pi _{n+1}^\bullet } \ar[r]^{\varphi _ n^\bullet } & {K_ n^\bullet } \ar[d]^{\pi _ n^\bullet } \\ {K_{n+1}^\bullet } \ar[r]_{\varphi _ n^\bullet } & {K_ n^\bullet .} } \]
However, this diagram does commute in the derived category, hence it commutes up to homotopy. We inductively replace $\pi _ n^\bullet $ for $n\geq 2$ by homotopic maps of complexes making these diagrams commute. Namely, if $h^ i : K_{n+1}^ i \to K_ n^{i-1}$ is a homotopy, i.e.,
\[ \pi _ n^\bullet \circ \varphi _ n^\bullet - \varphi _ n^\bullet \circ \pi _{n + 1}^\bullet = dh + hd, \]
then we choose $\tilde h^ i : K_{n+1}^ i\to K_{n+1}^{i-1}$ lifting $h^ i$. This is possible because $K_{n+1}^ i$ free and $K_{n+1}^{i-1}\to K_ n^{i-1}$ is surjective. Then replace $\pi _ n^\bullet $ by $\tilde\pi _ n^\bullet $ defined by
\[ \tilde\pi _{n+1}^\bullet = \pi _{n+1}^\bullet + d\tilde h+\tilde hd. \]
With this choice of $\{ \pi _ n^\bullet \} $, the above diagrams commute, and the maps fit together to define an endomorphism $\pi _\infty ^\bullet = \mathop{\mathrm{lim}}\nolimits _ n\pi _ n^\bullet $ of $K_\infty ^\bullet $. Then part i is clear: the elements $t_ n = \sum (-1)^ i \text{Tr}\left(\pi _ n^ i |_{K_ n^ i}\right)$ fit into an element $t_\infty $ of $\mathbf{Z}_\ell $. Moreover
\begin{align*} t_\infty & = \sum (-1)^ i \text{Tr}_{\mathbf{Z}_\ell }(\pi _\infty ^ i |_{K_\infty ^ i}) \\ & = \sum (-1)^ i \text{Tr}_{\mathbf{Q}_\ell }( \pi _\infty ^ i |_{K_\infty ^ i \otimes _{\mathbf{Z}_\ell }\mathbf{Q}_\ell }) \\ & = \sum (-1)^ i \text{Tr}( \pi _\infty |_{H^ i(K_\infty ^\bullet \otimes \mathbf{Q}_\ell )}) \end{align*}
where the last equality follows from the fact that $\mathbf{Q}_\ell $ is a field, so the complex $K_\infty ^\bullet \otimes \mathbf{Q}_\ell $ is quasi-isomorphic to its cohomology $H^ i(K_\infty ^\bullet \otimes \mathbf{Q}_\ell )$. The latter is also equal to $H^ i(K_\infty ^\bullet )\otimes _{\mathbf{Z}}\mathbf{Q}_\ell = H_\infty ^ i \otimes \mathbf{Q}_\ell $, which finishes the proof of the lemma, and also that of Theorem 64.20.5.
$\square$
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