## Tag `03UY`

## 50.101. Cohomological interpretation

This is how Grothendieck interpreted the $L$-function.

Theorem 50.101.1 (Finite Coefficients). Let $X$ be a scheme of finite type over a finite field $k$. Let $\Lambda$ be a finite ring of order prime to the characteristic of $k$ and $\mathcal{F}$ a constructible flat $\Lambda$-module on $X_{\acute{e}tale}$. Then $$ L(X, \mathcal{F}) = \det(1 - \pi_X^*~T |_{R\Gamma_c(X_{\bar k}, \mathcal{F})})^{-1} \in \Lambda[[T]]. $$

Proof.Omitted. $\square$Thus far, we don't even know whether each cohomology group $H^i_c(X_{\bar k}, \mathcal{F})$ is free.

Theorem 50.101.2 (Adic sheaves). Let $X$ be a scheme of finite type over a finite field $k$, and $\mathcal{F}$ a $\mathbf{Q}_\ell$-sheaf on $X$. Then $$ L(X, \mathcal{F}) = \prod\nolimits_i \det(1 - \pi_X^*T |_{H_c^i(X_{\bar k} , \mathcal{F})})^{(-1)^{i + 1}} \in \mathbf{Q}_\ell[[T]]. $$

Proof.This is sketched below. $\square$Remark 50.101.3. Since we have only developed some theory of traces and not of determinants, Theorem 50.101.1 is harder to prove than Theorem 50.101.2. We will only prove the latter, for the former see [SGA4.5]. Observe also that there is no version of this theorem more general for $\mathbf{Z}_\ell$ coefficients since there is no $\ell$-torsion.

We reduce the proof of Theorem 50.101.2 to a trace formula. Since $\mathbf{Q}_\ell$ has characteristic 0, it suffices to prove the equality after taking logarithmic derivatives. More precisely, we apply $T\frac{d}{dT} \log$ to both sides. We have on the one hand \begin{align*} T \frac{d}{dT} \log L(X, \mathcal{F}) & = T\frac{d}{dT} \log \prod_{x \in |X|} \det(1 - \pi_x^* T^{\deg x} |_{\mathcal{F}_{\bar x}})^{-1} \\ & = \sum_{x \in |X|} T \frac{d}{dT} \log ( \det(1 - \pi_x^* T^{\deg x} |_{\mathcal{F}_{\bar x}})^{-1}) \\ & = \sum_{x \in |X|} \deg x \sum_{n \geq 1} \text{Tr}((\pi_x^n)^* |_{\mathcal{F}_{\bar x}}) T^{n\deg x} \end{align*} where the last equality results from the formula $$ T\frac{d}{dT}\log\left(\det\left(1-fT|_M\right)^{-1}\right) = \sum_{n\geq 1} \text{Tr}(f^n|_M)T^n $$ which holds for any commutative ring $\Lambda$ and any endomorphism $f$ of a finite projective $\Lambda$-module $M$. On the other hand, we have \begin{align*} & T\frac{d}{dT} \log\left( \prod\nolimits_i \det(1-\pi_X^*T |_{H_c^i\left(X_{\bar k} , \mathcal{F}\right)})^{(-1)^{i+1}} \right) \\ & = \sum\nolimits_i (-1)^i \sum\nolimits_{n \geq 1} \text{Tr}\left((\pi_X^n)^* |_{H_c^i(X_{\bar k},\mathcal{F})}\right) T^n \end{align*} by the same formula again. Now, comparing powers of $T$ and using the Mobius inversion formula, we see that Theorem 50.101.2 is a consequence of the following equality $$ \sum_{d | n} d \sum_{x \in |X| \atop \deg x = d} \text{Tr}((\pi_X^{n/d})^* |_{\mathcal{F}_{\bar x}}) = \sum_i (-1)^i \text{Tr}((\pi^n_X)^* |_{H^i_c(X_{\bar k}, \mathcal{F})}). $$ Writing $k_n$ for the degree $n$ extension of $k$, $X_n = X \times_{\mathop{\rm Spec} k} \mathop{\rm Spec}(k_n)$ and $_n\mathcal{F} = \mathcal{F}|_{X_n}$, this boils down to $$ \sum_{x \in X_n(k_n)} \text{Tr}(\pi_X^* |_{_n\mathcal{F}_{\bar x}}) = \sum_i (-1)^i \text{Tr}((\pi^n_X)^* |_{H^i_c({(X_n)}_{\bar k}, _n\mathcal{F})}) $$ which is a consequence of Theorem 50.101.5.

Theorem 50.101.4. Let $X/k$ be as above, let $\Lambda$ be a finite ring with $\#\Lambda \in k^*$ and $K\in D_{ctf}(X, \Lambda)$. Then $R\Gamma_c(X_{\bar k}, K)\in D_{perf}(\Lambda)$ and $$ \sum_{x\in X(k)}\text{Tr}\left(\pi_x |_{K_{\bar x}}\right) = \text{Tr}\left(\pi_X^* |_{R\Gamma_c(X_{\bar k}, K )}\right). $$

Proof.Note that we have already proved this (REFERENCE) when $\dim X \leq 1$. The general case follows easily from that case together with the proper base change theorem. $\square$Theorem 50.101.5. Let $X$ be a separated scheme of finite type over a finite field $k$ and $\mathcal{F}$ be a $\mathbf{Q}_\ell$-sheaf on $X$. Then $\dim_{\mathbf{Q}_\ell}H_c^i(X_{\bar k}, \mathcal{F})$ is finite for all $i$, and is nonzero for $0\leq i \leq 2 \dim X$ only. Furthermore, we have $$ \sum_{x\in X(k)} \text{Tr}\left(\pi_x |_{\mathcal{F}_{\bar x}}\right) = \sum_i (-1)^i\text{Tr}\left(\pi_X^* |_{H_c^i(X_{\bar k}, \mathcal{F})}\right). $$

Proof.We explain how to deduce this from Theorem 50.101.4. We first use some étale cohomology arguments to reduce the proof to an algebraic statement which we subsequently prove.Let $\mathcal{F}$ be as in the theorem. We can write $\mathcal{F}$ as $\mathcal{F}'\otimes \mathbf{Q}_\ell$ where $\mathcal{F}' = \left\{\mathcal{F}'_n\right\}$ is a $\mathbf{Z}_\ell$-sheaf without torsion, i.e., $\ell : \mathcal{F}'\to \mathcal{F}'$ has trivial kernel in the category of $\mathbf{Z}_\ell$-sheaves. Then each $\mathcal{F}_n'$ is a flat constructible $\mathbf{Z}/\ell^n\mathbf{Z}$-module on $X_{\acute{e}tale}$, so $\mathcal{F}'_n \in D_{ctf}(X, \mathbf{Z}/\ell^n\mathbf{Z})$ and $\mathcal{F}_{n+1}' \otimes^{\mathbf{L}}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}} \mathbf{Z}/\ell^n\mathbf{Z} = \mathcal{F}_n'$. Note that the last equality holds also for standard (non-derived) tensor product, since $\mathcal{F}'_n$ is flat (it is the same equality). Therefore,

- the complex $K_n = R\Gamma_c\left(X_{\bar k}, \mathcal{F}_n'\right)$ is perfect, and it is endowed with an endomorphism $\pi_n : K_n\to K_n$ in $D(\mathbf{Z}/\ell^n\mathbf{Z})$,
- there are identifications $$ K_{n+1} \otimes^{\mathbf{L}}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}} \mathbf{Z}/\ell^n\mathbf{Z} = K_n $$ in $D_{perf}(\mathbf{Z}/\ell^n\mathbf{Z})$, compatible with the endomorphisms $\pi_{n+1}$ and $\pi_n$ (see \cite[Rapport 4.12]{SGA4.5}),
- the equality $\text{Tr}\left(\pi_X^* |_{K_n}\right) = \sum_{x\in X(k)} \text{Tr}\left(\pi_x |_{(\mathcal{F}'_n)_{\bar x}}\right)$ holds, and
- for each $x\in X(k)$, the elements $\text{Tr}(\pi_x |_{\mathcal{F}'_{n, \bar x}}) \in \mathbf{Z}/\ell^n\mathbf{Z}$ form an element of $\mathbf{Z}_\ell$ which is equal to $\text{Tr}(\pi_x |_{\mathcal{F}_{\bar x}}) \in \mathbf{Q}_\ell$.
It thus suffices to prove the following algebra lemma. $\square$

Lemma 50.101.6. Suppose we have $K_n\in D_{perf}(\mathbf{Z}/\ell^n\mathbf{Z})$, $\pi_n : K_n\to K_n$ and isomorphisms $\varphi_n : K_{n+1} \otimes^\mathbf{L}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}} \mathbf{Z}/\ell^n\mathbf{Z} \to K_n$ compatible with $\pi_{n+1}$ and $\pi_n$. Then

- the elements $t_n = \text{Tr}(\pi_n |_{K_n})\in \mathbf{Z}/\ell^n\mathbf{Z}$ form an element $t_\infty = \{t_n\}$ of $\mathbf{Z}_\ell$,
- the $\mathbf{Z}_\ell$-module $H_\infty^i = \mathop{\rm lim}\nolimits_n H^i(k_n)$ is finite and is nonzero for finitely many $i$ only, and
- the operators $H^i(\pi_n): H^i(K_n)\to H^i(K_n)$ are compatible and define $\pi_\infty^i : H_\infty^i\to H_\infty^i$ satisfying $$ \sum (-1)^i \text{Tr}( \pi_\infty^i |_{H_\infty^i \otimes_{\mathbf{Z}_\ell}\mathbf{Q}_\ell}) = t_\infty. $$

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 $K_\infty^i = \mathop{\rm lim}\nolimits_n \{K_n^i, \varphi_n^i\}$ 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 $H^i\left(K_\infty^\bullet\right) = \mathop{\rm lim}\nolimits_n H^i\left(K_n^\bullet\right)$ (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 $\pi_\infty^\bullet = \mathop{\rm lim}\nolimits_n\pi_n^\bullet$ of $K_\infty^\bullet$. Then partiis 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 50.101.5. $\square$

The code snippet corresponding to this tag is a part of the file `etale-cohomology.tex` and is located in lines 17481–17776 (see updates for more information).

```
\section{Cohomological interpretation}
\label{section-L-cohomological}
\noindent
This is how Grothendieck interpreted the $L$-function.
\begin{theorem}[Finite Coefficients]
\label{theorem-A}
Let $X$ be a scheme of finite type over a finite field $k$. Let $\Lambda$ be a
finite ring of order prime to the characteristic of $k$ and $\mathcal{F}$ a
constructible flat $\Lambda$-module on $X_\etale$. Then
$$
L(X, \mathcal{F}) =
\det(1 - \pi_X^*\ T |_{R\Gamma_c(X_{\bar k}, \mathcal{F})})^{-1}
\in \Lambda[[T]].
$$
\end{theorem}
\begin{proof}
Omitted.
\end{proof}
\noindent
Thus far, we don't even know whether each cohomology group
$H^i_c(X_{\bar k}, \mathcal{F})$ is free.
\begin{theorem}[Adic sheaves]
\label{theorem-B}
Let $X$ be a scheme of finite type over a finite field $k$, and $\mathcal{F}$ a
$\mathbf{Q}_\ell$-sheaf on $X$. Then
$$
L(X, \mathcal{F}) =
\prod\nolimits_i
\det(1 - \pi_X^*T |_{H_c^i(X_{\bar k} , \mathcal{F})})^{(-1)^{i + 1}}
\in \mathbf{Q}_\ell[[T]].
$$
\end{theorem}
\begin{proof}
This is sketched below.
\end{proof}
\begin{remark}
\label{remark-which-is-harder}
Since we have only developed some theory of traces and not of determinants,
Theorem \ref{theorem-A}
is harder to prove than
Theorem \ref{theorem-B}.
We will only prove the latter, for the former see \cite{SGA4.5}.
Observe also that there is no version of this theorem more general for
$\mathbf{Z}_\ell$ coefficients since there is no $\ell$-torsion.
\end{remark}
\noindent
We reduce the proof of Theorem \ref{theorem-B} to a trace formula. Since
$\mathbf{Q}_\ell$ has characteristic 0, it suffices to prove the equality after
taking logarithmic derivatives. More precisely, we apply $T\frac{d}{dT} \log$
to both sides. We have on the one hand
\begin{align*}
T \frac{d}{dT} \log L(X, \mathcal{F})
& =
T\frac{d}{dT} \log
\prod_{x \in |X|} \det(1 - \pi_x^* T^{\deg x} |_{\mathcal{F}_{\bar x}})^{-1} \\
& =
\sum_{x \in |X|} T \frac{d}{dT} \log
( \det(1 - \pi_x^* T^{\deg x} |_{\mathcal{F}_{\bar x}})^{-1}) \\
& =
\sum_{x \in |X|} \deg x
\sum_{n \geq 1} \text{Tr}((\pi_x^n)^* |_{\mathcal{F}_{\bar x}}) T^{n\deg x}
\end{align*}
where the last equality results from the formula
$$
T\frac{d}{dT}\log\left(\det\left(1-fT|_M\right)^{-1}\right) = \sum_{n\geq 1}
\text{Tr}(f^n|_M)T^n
$$
which holds for any commutative ring $\Lambda$ and any endomorphism $f$ of a
finite projective $\Lambda$-module $M$. On the other hand, we have
\begin{align*}
& T\frac{d}{dT} \log\left(
\prod\nolimits_i
\det(1-\pi_X^*T |_{H_c^i\left(X_{\bar k} , \mathcal{F}\right)})^{(-1)^{i+1}}
\right) \\
& =
\sum\nolimits_i (-1)^i \sum\nolimits_{n \geq 1}
\text{Tr}\left((\pi_X^n)^* |_{H_c^i(X_{\bar k},\mathcal{F})}\right) T^n
\end{align*}
by the same formula again. Now, comparing powers of $T$ and using the Mobius
inversion formula, we see that Theorem \ref{theorem-B} is a consequence of the
following equality
$$
\sum_{d | n} d \sum_{x \in |X| \atop \deg x = d}
\text{Tr}((\pi_X^{n/d})^* |_{\mathcal{F}_{\bar x}}) =
\sum_i (-1)^i \text{Tr}((\pi^n_X)^* |_{H^i_c(X_{\bar k}, \mathcal{F})}).
$$
Writing $k_n$ for the degree $n$ extension of $k$,
$X_n = X \times_{\Spec k} \Spec(k_n)$ and
$_n\mathcal{F} = \mathcal{F}|_{X_n}$, this boils down
to
$$
\sum_{x \in X_n(k_n)} \text{Tr}(\pi_X^* |_{_n\mathcal{F}_{\bar x}})
=
\sum_i (-1)^i \text{Tr}((\pi^n_X)^* |_{H^i_c({(X_n)}_{\bar k}, _n\mathcal{F})})
$$
which is a consequence of Theorem \ref{theorem-C}.
%11.19.09
\begin{theorem}
\label{theorem-D}
Let $X/k$ be as above, let $\Lambda$ be a finite ring with $\#\Lambda \in k^*$
and $K\in D_{ctf}(X, \Lambda)$. Then $R\Gamma_c(X_{\bar k}, K)\in
D_{perf}(\Lambda)$ and
$$
\sum_{x\in X(k)}\text{Tr}\left(\pi_x |_{K_{\bar x}}\right) =
\text{Tr}\left(\pi_X^* |_{R\Gamma_c(X_{\bar k}, K )}\right).
$$
\end{theorem}
\begin{proof}
Note that we have already proved this (REFERENCE) when $\dim X \leq 1$. The
general case follows easily from that case together with the proper base change
theorem.
\end{proof}
\begin{theorem}
\label{theorem-C}
Let $X$ be a separated scheme of finite type over a finite field $k$ and
$\mathcal{F}$ be a $\mathbf{Q}_\ell$-sheaf on $X$. Then
$\dim_{\mathbf{Q}_\ell}H_c^i(X_{\bar k}, \mathcal{F})$ is finite for all $i$,
and is nonzero for $0\leq i \leq 2 \dim X$ only. Furthermore, we have
$$
\sum_{x\in X(k)} \text{Tr}\left(\pi_x |_{\mathcal{F}_{\bar x}}\right) =
\sum_i (-1)^i\text{Tr}\left(\pi_X^* |_{H_c^i(X_{\bar k}, \mathcal{F})}\right).
$$
\end{theorem}
\begin{proof}
We explain how to deduce this from Theorem \ref{theorem-D}.
We first use some \'etale cohomology arguments to reduce the proof
to an algebraic statement which we subsequently prove.
\medskip\noindent
Let $\mathcal{F}$ be as in the theorem. We can write
$\mathcal{F}$ as
$\mathcal{F}'\otimes \mathbf{Q}_\ell$ where $\mathcal{F}' =
\left\{\mathcal{F}'_n\right\}$ is a $\mathbf{Z}_\ell$-sheaf without torsion,
i.e., $\ell : \mathcal{F}'\to \mathcal{F}'$ has trivial kernel in the
category of $\mathbf{Z}_\ell$-sheaves. Then each $\mathcal{F}_n'$ is a flat
constructible $\mathbf{Z}/\ell^n\mathbf{Z}$-module on $X_\etale$, so
$\mathcal{F}'_n \in D_{ctf}(X, \mathbf{Z}/\ell^n\mathbf{Z})$ and
$\mathcal{F}_{n+1}'
\otimes^{\mathbf{L}}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}}
\mathbf{Z}/\ell^n\mathbf{Z} = \mathcal{F}_n'$.
Note that the last equality holds also
for standard (non-derived) tensor product, since $\mathcal{F}'_n$ is flat
(it is the same equality). Therefore,
\begin{enumerate}
\item
the complex $K_n = R\Gamma_c\left(X_{\bar k}, \mathcal{F}_n'\right)$ is perfect,
and it is endowed with an endomorphism $\pi_n : K_n\to K_n$ in
$D(\mathbf{Z}/\ell^n\mathbf{Z})$,
\item
there are identifications
$$
K_{n+1}
\otimes^{\mathbf{L}}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}}
\mathbf{Z}/\ell^n\mathbf{Z}
=
K_n
$$
in $D_{perf}(\mathbf{Z}/\ell^n\mathbf{Z})$, compatible with the endomorphisms
$\pi_{n+1}$ and $\pi_n$ (see \cite[Rapport 4.12]{SGA4.5}),
\item
the equality $\text{Tr}\left(\pi_X^* |_{K_n}\right) =
\sum_{x\in X(k)} \text{Tr}\left(\pi_x |_{(\mathcal{F}'_n)_{\bar x}}\right)$
holds, and
\item
for each $x\in X(k)$, the elements
$\text{Tr}(\pi_x |_{\mathcal{F}'_{n, \bar x}}) \in \mathbf{Z}/\ell^n\mathbf{Z}$
form an element of
$\mathbf{Z}_\ell$ which is equal to
$\text{Tr}(\pi_x |_{\mathcal{F}_{\bar x}}) \in \mathbf{Q}_\ell$.
\end{enumerate}
It thus suffices to prove the following algebra lemma.
\end{proof}
\begin{lemma}
\label{lemma-piece-together}
Suppose we have
$K_n\in D_{perf}(\mathbf{Z}/\ell^n\mathbf{Z})$, $\pi_n : K_n\to K_n$
and isomorphisms
$\varphi_n :
K_{n+1} \otimes^\mathbf{L}_{\mathbf{Z}/\ell^{n+1}\mathbf{Z}}
\mathbf{Z}/\ell^n\mathbf{Z}
\to K_n$
compatible with $\pi_{n+1}$ and $\pi_n$. Then
\begin{enumerate}
\item
the elements $t_n = \text{Tr}(\pi_n |_{K_n})\in \mathbf{Z}/\ell^n\mathbf{Z}$
form an element $t_\infty = \{t_n\}$ of $\mathbf{Z}_\ell$,
\item
the $\mathbf{Z}_\ell$-module $H_\infty^i = \lim_n H^i(k_n)$ is finite and
is nonzero for finitely many $i$ only, and
\item
the operators $H^i(\pi_n): H^i(K_n)\to H^i(K_n)$ are compatible and define
$\pi_\infty^i : H_\infty^i\to H_\infty^i$ satisfying
$$
\sum (-1)^i \text{Tr}(
\pi_\infty^i |_{H_\infty^i \otimes_{\mathbf{Z}_\ell}\mathbf{Q}_\ell}) =
t_\infty.
$$
\end{enumerate}
\end{lemma}
\begin{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
\begin{itemize}
\item
there exist $a, b\in \mathbf{Z}$ independent of $n$ such that $K_n^i = 0$ for
all $i\notin[a, b]$, and
\item
the rank of $K_n^i$ is independent of $n$.
\end{itemize}
Therefore, the module $K_\infty^i = \lim_n \{K_n^i, \varphi_n^i\}$ 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 $H^i\left(K_\infty^\bullet\right) = \lim_n
H^i\left(K_n^\bullet\right)$ (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 {\it 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 $\pi_\infty^\bullet =
\lim_n\pi_n^\bullet$ of $K_\infty^\bullet$. Then part {\it 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 \ref{theorem-C}.
\end{proof}
```

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