We have two goals:
- define étale motivic cohomology and related different definitions,
- study algebraic singular cohomology and relate it to étale cohomology.
Recall we have $\mathbb{Z}(q) = C_\ast(\mathbb{G}_ m^{\wedge q}[-q])$. We are also allowed to change coefficients.
Definition 1. We define $H_L^{p,q}(X, A) = \mathbb{H}_ \mathrm{et}^p(X, A(q) \vert_{X_\mathrm{et}})$.
The letter L stands for Lichtenbaum.
Example 2. When $q = 0$ we recover $H_L^{p,0}(X, A) = H_\mathrm{et}^p(X, A)$.
Let $\mathsf{D}$ be the derived category of étale sheaves with transfer. We had a further localization $\mathsf{DM}_ \mathrm{et}$, the category of étale motives, which was obtained from $\mathsf{D}$ by inverting $\mathbb{A}^1$-weak equivalences. So another possible definition is to compute the Ext groups from $\mathbb{Z}_ \mathrm{tr}(X)$ to $\mathbb{Z}(q)$.
Theorem 3. Over $\mathbb{Z}/n\mathbb{Z}$ where $n \in k^\times$, we have an isomorphism $$ H_L^{p,q}(X, \mathbb{Z}/n\mathbb{Z}) \cong \Hom_{\mathsf{DM}_ \mathrm{et}}(\mathbb{Z}_ \mathrm{tr}(X), \mathbb{Z}/n\mathbb{Z}(q)[p]). $$
Remark 4. When $\operatorname{char}(k) = \ell$ we have $\mathbb{Z}/\ell\mathbb{Z} \sim 0$. But the left hand side is nonzero.
Example 5. When $q = 1$, we have $\mu_n \sim \mathbb{Z}/n\mathbb{Z}(1)$. It follows that $H_L^{p,1}(X, \mathbb{Z}/n\mathbb{Z}) = H_\mathrm{et}^p(X, \mu_n)$.
In fact, this has a generalization.
Theorem 6. Assuming $n \in k^\times$, we have $H_L^{p,q}(X, \mathbb{Z}/n\mathbb{Z}) = H_\mathrm{et}^p(X, \mu_n^{\otimes q})$.
To show this, it now remains to show that the map $$ \mu_n^{\otimes q} \sim (\mathbb{Z}/n\mathbb{Z}(1))^{\otimes q} \to \mathbb{Z}/n\mathbb{Z}(q) $$ is a quasi-isomorphism. To see this, we first note that both sides are $\mathbb{A}^1$-local; the left hand side is locally constant, and the right hand side can also be checked. We will later check that $\alpha$ is a weak equivalence, and it follows that $\alpha$ is a quasi-isomorphism.
Remark 7. There are complicates pertaining to tensor products. There is a way to extend the tensor product $\mathbb{Z}_ \mathrm{tr}(X) \otimes_\mathrm{tr} \mathbb{Z}_ \mathrm{tr}(Y) = \mathbb{Z}_ \mathrm{tr}(X \times Y)$ to the derived category. There is also a derived tensor product as sheaves. But we will mostly sweep this under the rug. It turns out that the natural map $R(m) \otimes R(n) \to R(m+n)$ factors as $$ R(m) \otimes R(n) \to R(m) \otimes_\mathrm{tr} R(n) \to R(m+n). $$
Proof of the main results
Proposition 8. The map $\mu_n^{\otimes q} \to \mathbb{Z}/n\mathbb{Z}(q)$ is an $\mathbb{A}^1$-weak equivalence.
It turns out we have isomorphisms $$ \mu_n^{\otimes q} \cong \mu_n^{\otimes_\mathrm{tr} q} \cong \mu_n^{\otimes_\mathrm{tr}^L q}. $$ Because we have $\mu_n \cong \mathbb{Z}/n\mathbb{Z}(1)$, it now remains to show that $$ \mathbb{Z}/n\mathbb{Z}(1)^{\otimes_\mathrm{tr}^L q} \to \mathbb{Z}/n\mathbb{Z}(1)^{\otimes_\mathrm{tr} q} \to \mathbb{Z}/n\mathbb{Z}(q) $$ is a $\mathbb{A}^1$-weak equivalence. To show this, we recall that the singular chain complex is $\mathbb{A}^1$-equivalent to the original object, so we can replace this with $$ (\mathbb{Z}/n\mathbb{Z})_ \mathrm{tr}(\mathbb{G}_ m)[-1]^{\otimes_\mathrm{tr}^L q} \to ((\mathbb{Z}/n\mathbb{Z})_ \mathrm{tr}(\mathbb{G}_ m))^{\otimes_\mathrm{tr} q} [-q] \to (\mathbb{Z}/n\mathbb{Z})_ \mathrm{tr} (\mathbb{G}_ m^{\wedge q})[-q]. $$ The first map can be checked to be an weak equivalence, and the second map is an isomorphism by construction of $\otimes_\mathrm{tr}$, because we are extending from representable objects.
Proposition 9. If $n \in k^\times$ then we have $H_L^{p,q}(X, \mathbb{Z}/n\mathbb{Z}) \cong \Hom_{\mathsf{DM}_ \mathrm{et}}(\mathbb{Z}_ \mathrm{tr}(X), \mathbb{Z}/n\mathbb{Z}(q)[p])$.
Because both sides are $\mathbb{A}^1$-local, maps in $\mathsf{DM}_ \mathrm{et}$ can be understood as maps in $\mathsf{D}$. Then by the comparison theorem from last talk, the right hand side can be rewritten as $$ \Hom_{\mathsf{D}^-}(\mathbb{Z}_ \mathrm{tr}(X), \mathbb{Z}/n\mathbb{Z}(q)[p]) = \Ext^p(\mathbb{Z}_ \mathrm{tr}(X), \mathbb{Z}/n\mathbb{Z}(q)). $$ On the other hand, we saw previously that this can be identified with $\mathbb{H}_ \mathrm{et}^p(X, \mathbb{Z}/n\mathbb{Z}(q))$.
Algebraic singular cohomology
Definition 10. We define for $R$ a commutative ring $$ H_p^\mathrm{sing}(X, R) = H_p(C_\ast R_\mathrm{tr}(X)(\Spec k)). $$
Here is a classical theorem.
Theorem 11. Let $k$ be separably closed and $X/k$ be smooth, and let $\ell$ be prime invertible in $k$. Then $$ H_p^\mathrm{sing}(X, \mathbb{Z}/\ell\mathbb{Z})^\vee = H_\mathrm{et}^p(X, \mathbb{Z}/\ell\mathbb{Z}). $$
Here is the proof of this theorem. Let $\mathsf{D}_ R^-$ be the derived category of étale sheaves with transfer valued in $R$-modules. Then we have a diagram $$ \begin{CD} \Hom_{\mathsf{D}_ R^-}(C_\ast R_\mathrm{tr}(X), R[n]) @>>> \Hom_R(H_n^\mathrm{sing}(X, R), R) \br @VVV \br \Hom_{\mathsf{D}_ R^-}(R_\mathrm{tr}(X), R[n]) @>>> H_\mathrm{et}^n(X, R). \end{CD} $$ The bottom map is an isomorphism as we have seen before. The left map is an isomorphism because $R[n]$ is $\mathbb{A}^1$-local and the map between the sources is an $\mathbb{A}^1$-weak equivalence. It remains to see that the top map is an isomorphism.
Lemma 12. Let $k$ be separably closed, and let $C$ be a bounded above complex of étale sheaves of $R$-modules with transfer. Assume that the sheaves $H_n C$ are locally constant and projective. Then we have an isomorphism $$ \Hom_{\mathsf{D}_ R^-}(C, R[n]) = \Hom_R(H^n(C)(\Spec k), R). $$
To see this, we take an injective resolution $R \to \mathscr{I}^\bullet$. Then we have $\mathbb{R}\Hom(C, R) = \Hom(C, I)$ as complexes, and so we have a spectral sequence $$ E_2^{p,q} = \Ext^p(H_q C, R) \Longrightarrow H^{p+q} \mathbb{R}\Hom(C, R). $$ Because $H_q C$ are all projective, we actually have $$ \Ext^0(H_n C, R) \cong \Hom_{\mathsf{D}_ R^-}(C, R[n]). $$ On the other hand, $P = H^n C$ is a summand of $R^{\oplus \alpha}$ for some $\alpha$. Then we have $$ \Ext^n(P, R) \hookrightarrow \Ext^n(R^{\oplus \alpha}, R) = \prod_\alpha \Ext^n(R, R), $$ where we can compute this Ext term as $$ \Ext^n(R, R) = \Ext(R_\mathrm{tr}(\Spec k), R) = H_\mathrm{et}^n(\Spec k, R). $$ This vanishes for $n \gt 0$ and for $n = 0$ this is $R$ itself.
Remark 13. Here, $\ell$ need not be prime. If we consider a short exact sequence of sheaves $$ 0 \to \mathbb{Z}/\ell_1 \mathbb{Z} \to \mathbb{Z}/\ell_1 \ell_2 \mathbb{Z} \to \mathbb{Z}/\ell_2 \mathbb{Z} \to 0 $$ then we can use the five lemma to deduce the same fact for products of primes.