We recall some notations and definitions. Let $\mathsf{Cor}_ k$ be the category of correspondences over a field $k$. We defined a presheaf of transfer as a contravariant functor $\mathsf{Cor}_ k \to \mathsf{Ab}$, and denoted the category of those as $\mathsf{PST}(k)$. For $X$ a smooth variety we also defined $$ \mathbb{Z}_ \mathrm{tr}(X) \colon U \mapsto \operatorname{Cor}(U, X). $$ It is clear that $\mathbb{Z}_ \mathrm{tr}(\Spec k) = \mathbb{Z}$ as presheaves. Given a pointed variety $(X, x)$ we also defined $$ \mathbb{Z}_ \mathrm{tr}(X, x) = \coker(\mathbb{Z} \to \mathbb{Z}_ \mathrm{tr}(X)). $$ This has a natural splitting $$ \mathbb{Z}_ \mathrm{tr}(X) \cong \mathbb{Z} \oplus \mathbb{Z}_ \mathrm{tr}(X, x). $$
Definition 1. For pointed schemes $(X_i, x_i)$, we also define $$ \mathbb{Z}_ \mathrm{tr}(\wedge(X_i, x_i)) = \coker\biggl( \bigoplus_i \mathbb{Z}_ \mathrm{tr}(X_1 \times \dotsb \times \hat{X}_ i \times \dotsb \times X_n) \to \mathbb{Z}_ \mathrm{tr}(X_1 \times \dotsb \times X_n) \biggr). $$
This can also be thought of as the cohomology of the last term in the complex $$ 0 \to \mathbb{Z} \to \bigoplus_i \mathbb{Z}_ \mathrm{tr}(X_i) \to \bigoplus_{i,j} \mathbb{Z}_ \mathrm{tr}(X_i \times X_j) \to \dotsb \to \mathbb{Z}_ \mathrm{tr}(X_1 \times \dotsb \times X_n) \to 0. $$
Definition 2. For $\mathscr{F} \in \mathsf{PST}(k)$ we defined a chain complex $$ C_\ast \mathscr{F} = [\dotsb \to \mathscr{F}(U \times \Delta^2) \to \mathscr{F}(U \times \Delta^1) \to \mathscr{F}(U) \to 0]. $$
Definition 3. For $q \in \mathbb{Z}_ {\ge 0}$ an integer, we define the complex $$ \mathbb{Z}(q) = C_\ast \mathbb{Z}_ \mathrm{tr}(\mathbb{G}_ m^{\wedge q})[-q] $$ where $\mathbb{G}_ m = (\mathbb{A}^1 - \lbrace 0 \rbrace, 1)$.
When $q = 0$ we have $$ \mathbb{Z}(0) = C_\ast \mathbb{Z} \simeq \mathbb{Z} $$ and when $q = 1$ we have $$ \mathbb{Z}(1) = C_\ast \mathbb{Z}_ \mathrm{tr}(\mathbb{G}_ m)[-1]. $$
Proposition 4. For every variety $X$, the presheaf $\mathbb{Z}_ \mathrm{tr}(X)$ is a sheaf in the Zariski topology.
Proof.
Suppose we have $U = U_1 \cup U_2$, and now we need to show that $$ 0 \to \operatorname{Cor}(U, X) \to \operatorname{Cor}(U_1, X) \oplus \operatorname{Cor}(U_2, X) \to \operatorname{Cor}(U_1 \cap U_2, X) $$ is exact. If we write $Y = \sum n_i Y_i \in \operatorname{Cor}(U_1, X)$ and $Z = \sum m_j, Z_j \in \operatorname{Cor}(U_2, X)$. Now we have $Y \vert_{U_1 \cap U_2} = Z \vert_{U_1 \cap U_2}$, so we can now match $Y_i$ with $Z_j$ and glue them together.
Motivic cohomology
Definition 5. For $X$ smooth, we define the motivic cohomology as $$ H^{p,q}(X, \mathbb{Z}) = H_\mathrm{Zar}^p(X, \mathbb{Z}(q)), $$ and similarly $$ H^{p,q}(X, A) = H_\mathrm{Zar}^p(X, A(q)), \quad A(q) = \mathbb{Z}(q) \otimes A. $$
Here, $H_\mathrm{Zar}^p$ means hypercohomology, where it is regarded as a complex of sheaves on $X_\mathrm{Zar}$.
Proposition 6. We have $H^{p,q}(X, A) = 0$ when $p \gt q + \dim X$.
Proof.
Recall we have by definition $$ \mathbb{Z}(q)^i = C_{q-i} \mathbb{Z}_ \mathrm{tr}(\mathbb{G}_ m^{\wedge q}) $$ and $i \gt q$. Now when we take hypercohomology the degree increases by at most $\dim X$.
Theorem 7. For $X$ smooth over $k$ perfect, we have an isomorphism $$ H^{p,q}(X) \cong \mathrm{CH}^q(X, 2q-p). $$ Here, for $p = 2q$ we have the classical Chow groups.
So this will imply a strong vanishing result, which is that when $p \gt 2q$ then this also vanishes. There is also a nice property that rationally these cohomology $H^{p,q}$ doesn’t change when we replace $k$ with a finite separable extension. We will be able to see this at least when $p = 1$.
The sheaf $\mathbb{Z}(1)$
Theorem 8. There is a quasi-isomorphism $$ \mathbb{Z}(1) \simeq \mathscr{O}^\times[-1], $$ and hence $H^{1,1}(X) = \mathscr{O}^\ast(X)$ and $H^{1,2}(X) = \operatorname{Pic}(X)$.
We consider the functor $M \colon \mathsf{Sm} / k \to \mathsf{Ab}$ sending $X$ to the group of rational functions on $X \times \mathbb{P}^1$ that are regular on a neighborhood of $X \times \lbrace 0, \infty \rbrace$, and moreover take the value $1$ on $X \times \lbrace 0, \infty \rbrace$.
Lemma 9. For $f \in M(X)$, the Weil divisor $\operatorname{Div}(f)$ may be regarded as a correspondence in $\operatorname{Cor}(X, \mathbb{A}^1 - \lbrace 0 \rbrace)$.
Proof.
We just need to show that $\operatorname{Div}(f)$ is a difference of finite schemes over $X$. We can work locally, and then we can write down the function $f$ as a ratio of two monic polynomials of the same degree.
Lemma 10. For connected $X$, there is a short exact sequence $$ 0 \to M(X) \to \mathbb{Z}_ \mathrm{tr}(\mathbb{A}^1 - \lbrace 0 \rbrace)(X) \xrightarrow{\lambda} \mathbb{Z} \oplus \mathscr{O}^\ast(X) \to 0. $$
Proof.
How do we construct this $\lambda$? The claim is that for all $Z \in \operatorname{Cor}(X, \mathbb{A}^1 - \lbrace 0 \rbrace)$ there exists a unique rational function $f$ on $X \times \mathbb{P}^1$ and an integer $n$ such that
- $f/t^n$ is $1$ on $X \times \infty$,
- $f$ is invertible on $X \times 0$,
- $\operatorname{Div}(f) = Z$.
Now we define $\lambda(f) = (n, (-1)^n f(0))$ and its kernel is $M(X)$ by definition.
It follows that $M$ is also a presheaf with transfer, and we have a short exact sequence of complexes $$ 0 \to C_\ast M \to C_\ast \mathbb{Z}_ \mathrm{tr}(\mathbb{A}^1 - \lbrace 0 \rbrace) \to C_\ast(\mathbb{Z} \oplus \mathscr{O}^\times) \to 0. $$ On the other hand, recall that $\mathbb{Z}_ \mathrm{tr}(\mathbb{A}^1 - \lbrace 0 \rbrace) = \mathbb{Z} \oplus \mathbb{Z}_ \mathrm{tr}(\mathbb{G}_ m)$ and also $C_\ast \mathbb{Z} = \mathbb{Z}$ and $C_\ast(\mathscr{O}^\times) \simeq \mathscr{O}^\ast$. It follows that $$ 0 \to C_\ast M \to \mathbb{Z}(1)[1] \to \mathscr{O}^\ast \to 0 $$ is short exact.
Claim 11. The chain complex $C_\ast M$ is acyclic.
Proof.
Let $f \in C_i M(X)$ be a cycle, i.e., it vanishes in $C_{i-1} M(X)$. This means that the rational function $f$ on $X \times \Delta^i \times \mathbb{P}^1$ such that
- it takes the value $1$ on $Z = X \times \Delta^i \times \lbrace 0, \infty \rbrace$,
- it takes the value $1$ on all the faces $X \times \Delta^{i-1} \times \mathbb{P}^1$.
Then we can find an $\mathbb{A}^1$-homotopy $f \sim 1$ by using the formula $1 - t (1-f) = h_t(f)$.