**Proof of Theorem 64.16.1.**
The proof proceeds in a number of steps.

Step 1. *Let $j : \mathcal{U}\hookrightarrow X$ be an open immersion with complement $Y = X - \mathcal{U}$ and $i : Y \hookrightarrow X$. Then $T''(X, K) = T''(\mathcal{U}, j^{-1} K)+ T''(Y, i^{-1}K)$ and $T'(X, K) = T'(\mathcal{U}, j^{-1} K)+ T'(Y, i^{-1}K)$.*

This is clear for $T'$. For $T''$ use the exact sequence

\[ 0\to j_!j^{-1} K \to K \to i_* i^{-1} K \to 0 \]

to get a filtration on $K$. This gives rise to an object $\widetilde K \in DF(X, \Lambda )$ whose graded pieces are $j_!j^{-1}K$ and $i_*i^{-1}K$, both of which lie in $D_{ctf}(X, \Lambda )$. Then, by filtered derived abstract nonsense (INSERT REFERENCE), $R\Gamma _ c(X_{\bar k}, K)\in DF_{perf}(\Lambda )$, and it comes equipped with $\pi _ x^*$ in $DF_{perf}(\Lambda )$. By the discussion of traces on filtered complexes (INSERT REFERENCE) we get

\begin{eqnarray*} \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, K)}) & = & \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, j_!j^{-1}K)}) + \text{Tr}(\pi _ X^* |_{R\Gamma _ c(X_{\bar k}, i_*i^{-1}K)}) \\ & = & T”(U, i^{-1}K) + T”(Y, i^{-1}K). \end{eqnarray*}

Step 2. *The theorem holds if $\dim X\leq 0$. *

Indeed, in that case

\[ R\Gamma _ c(X_{\bar k}, K) = R\Gamma (X_{\bar k}, K) = \Gamma (X_{\bar k}, K) = \bigoplus \nolimits _{\bar x \in X_{\bar k}} K_{\bar x} \]

which comes equipped with the endormophism $\pi _ X^*$. Since the fixed points of $\pi _ X : X_{\bar k}\to X_{\bar k}$ are exactly the points $\bar x \in X_{\bar k}$ which lie over a $k$-rational point $x \in X(k)$ we get

\[ \text{Tr}\big (\pi _ X^*|_{R\Gamma _ c(X_{\bar k}, K)}\big ) = \sum \nolimits _{x \in X(k)} \text{Tr}(\pi _ x|_{K_{\bar x}}). \]

as desired.

Step 3. *It suffices to prove the equality $T'(\mathcal{U}, \mathcal{F}) = T''(\mathcal{U}, \mathcal{F})$ in the case where *

*
*
$\mathcal{U}$ is a smooth irreducible affine curve over $k$,

$\mathcal{U}(k) = \emptyset $,

$K=\mathcal{F}$ is a finite locally constant sheaf of $\Lambda $-modules on $\mathcal{U}$ whose stalk(s) are finite projective $\Lambda $-modules, and

$\Lambda $ is killed by a power of a prime $\ell $ and $\ell \in k^*$.

* *
Indeed, because of Step 2, we can throw out any finite set of points. But we have only finitely many rational points, so we may assume there are none^{1}. We may assume that $\mathcal{U}$ is smooth irreducible and affine by passing to irreducible components and throwing away the bad points if necessary. The assumptions of $\mathcal{F}$ come from unwinding the definition of $D_{ctf}(X, \Lambda )$ and those on $\Lambda $ from considering its primary decomposition.

For the remainder of the proof, we consider the situation

\[ \xymatrix{ \mathcal{V} \ar[d]_ f \ar[r] & Y \ar[d]^{\bar f} \\ \mathcal{U} \ar[r] & X } \]

where $\mathcal{U}$ is as above, $f$ is a finite étale Galois covering, $\mathcal{V}$ is connected and the horizontal arrows are projective completions. Denoting $G=\text{Aut}(\mathcal{V}|\mathcal{U})$, we also assume (as we may) that $f^{-1}\mathcal{F} =\underline M$ is constant, where the module $M = \Gamma (\mathcal{V}, f^{-1}\mathcal{F})$ is a $\Lambda [G]$-module which is finite and projective over $\Lambda $. This corresponds to the trivial monoid extension

\[ 1\to G\to \Gamma = G \times \mathbf{N}\to \mathbf{N}\to 1. \]

In that context, using the reductions above, we need to show that $T''(\mathcal{U}, \mathcal{F}) = 0$.

Step 4. *There is a natural action of $G$ on $f_*f^{-1}\mathcal{F}$ and the trace map $f_*f^{-1}\mathcal{F}\to \mathcal{F}$ defines an isomorphism*

\[ (f_*f^{-1}\mathcal{F})\otimes _{\Lambda [G]} \Lambda = (f_*f^{-1}\mathcal{F})_ G \cong \mathcal{F}. \]

To prove this, simply unwind everything at a geometric point.

Step 5. *Let $A = \mathbf{Z}/\ell ^ n \mathbf{Z}$ with $n\gg 0$. Then $f_*f^{-1}\mathcal{F} \cong (f_*\underline A) \otimes _{\underline A} \underline M$ with diagonal $G$-action.*

Step 6. *There is a canonical isomorphism $(f_*\underline A \otimes _{\underline A} \underline M) \otimes _{\Lambda [G]} \underline\Lambda \cong \mathcal{F}$. *

In fact, this is a derived tensor product, because of the projectivity assumption on $\mathcal{F}$.

Step 7. *There is a canonical isomorphism *

*
\[ R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F}) = (R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)\otimes _ A^\mathbf {L} M)\otimes _{\Lambda [G]}^\mathbf {L} \Lambda , \]
*

* compatible with the action of $\pi ^*_\mathcal {U}$. *
This comes from the universal coefficient theorem, i.e., the fact that $R\Gamma _ c$ commutes with $\otimes ^\mathbf {L}$, and the flatness of $\mathcal{F}$ as a $\Lambda $-module.

We have

\begin{eqnarray*} \text{Tr}( \pi _\mathcal {U}^* |_{R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F})}) & = & {\sum _{g \in G}}' \text{Tr}_{\Lambda }^{Z_ g} \left( (g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)\otimes _ A^\mathbf {L} M} \right) \\ & = & {\sum _{g\in G}}' \text{Tr}_ A^{Z_ g} ( (g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)} ) \cdot \text{Tr}_\Lambda (g|_ M) \end{eqnarray*}

where $\Gamma $ acts on $R\Gamma _ c(\mathcal{U}_{\bar k}, \mathcal{F})$ by $G$ and $(e, 1)$ acts via $\pi _\mathcal {U}^*$. So the monoidal extension is given by $\Gamma = G \times \mathbf{N} \to \mathbf{N}$, $\gamma \mapsto 1$. The first equality follows from Lemma 64.15.9 and the second from Lemma 64.15.8.

Step 8. *It suffices to show that $\text{Tr}_ A^{Z_ g}((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)}) \in A$ maps to zero in $\Lambda $. *

Recall that

\begin{eqnarray*} \# Z_ g \cdot \text{Tr}_ A^{Z_ g}((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)}) & = & \text{Tr}_ A((g, \pi _\mathcal {U}^*) |_{R\Gamma _ c(\mathcal{U}_{\bar k}, f_*A)})\\ & = & \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k}, A)}). \end{eqnarray*}

The first equality is Lemma 64.15.7, the second is the Leray spectral sequence, using the finiteness of $f$ and the fact that we are only taking traces over $A$. Now since $A=\mathbf{Z}/\ell ^ n\mathbf{Z}$ with $n \gg 0$ and $\# Z_ g = \ell ^ a$ for some (fixed) $a$, it suffices to show the following result.

Step 9. *We have $\text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}, A)}) = 0$ in $A$.*

By additivity again, we have

\begin{eqnarray*} & \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k} A)}) + \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(Y-\mathcal{V})_{\bar k}, A)}) \\ & = \text{Tr}_ A((g^{-1}\pi _ Y)^* |_{R\Gamma (Y_{\bar k}, A)}) \end{eqnarray*}

The latter trace is the number of fixed points of $g^{-1}\pi _ Y$ on $Y$, by Weil's trace formula Theorem 64.14.4. Moreover, by the 0-dimensional case already proven in step 2,

\[ \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^*|_{R\Gamma _ c(Y-\mathcal{V})_{\bar k}, A)}) \]

is the number of fixed points of $g^{-1}\pi _ Y$ on $(Y-\mathcal{V})_{\bar k}$. Therefore,

\[ \text{Tr}_ A((g^{-1}\pi _\mathcal {V})^* |_{R\Gamma _ c(\mathcal{V}_{\bar k}, A)}) \]

is the number of fixed points of $g^{-1}\pi _ Y$ on $\mathcal{V}_{\bar k}$. But there are no such points: if $\bar y\in Y_{\bar k}$ is fixed under $g^{-1}\pi _ Y$, then $\bar f(\bar y) \in X_{\bar k}$ is fixed under $\pi _ X$. But $\mathcal{U}$ has no $k$-rational point, so we must have $\bar f(\bar y)\in (X-\mathcal{U})_{\bar k}$ and so $\bar y\notin \mathcal{V}_{\bar k}$, a contradiction. This finishes the proof.
$\square$

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