Remark 42.32.8. Let $(S, \delta )$ be as in Situation 42.7.1. Let $Y$ be locally of finite type over $S$. Let $r \geq 0$. Let $f : X \to Y$ be a morphism of schemes. Assume every $y \in Y$ is contained in an open $V \subset Y$ such that $f^{-1}(V) \cong V \times \mathbf{A}^ r$ as schemes over $V$. In this remark we sketch a proof of the fact that $f^* : \mathop{\mathrm{CH}}\nolimits _ k(Y) \to \mathop{\mathrm{CH}}\nolimits _ k(X)$ is an isomorphism. First, by Lemma 42.32.1 the map is surjective. Let $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Y)$ with $f^*\alpha = 0$. We will prove that $\alpha = 0$.

Step 1. We may assume that $\dim _\delta (Y) < \infty$. (This is immediate in all cases in practice so we suggest the reader skip this step.) Namely, any rational equivalence witnessing that $f^*\alpha = 0$ on $X$, will use a locally finite collection of integral closed subschemes of dimension $k + r + 1$. Taking the union of the closures of the images of these in $Y$ we get a closed subscheme $Y' \subset Y$ of $\dim _\delta (Y') \leq k + r + 1$ such that $\alpha$ is the image of some $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ k(Y')$ and such that $(f')^*\alpha = 0$ where $f'$ is the base change of $f$ to $Y'$.

Step 2. Assume $d = \dim _\delta (Y) < \infty$. Then we can use induction on $d$. If $d < k$, then $\alpha = 0$ and we are done; this is the base case of the induction. In general, our assumption on $f$ shows we can choose a dense open $V \subset Y$ such that $U = f^{-1}(V) = \mathbf{A}^ r_ V$. Denote $Y' \subset Y$ the complement of $V$ as a reduced closed subscheme and set $X' = f^{-1}(Y')$. Consider

$\xymatrix{ \mathop{\mathrm{CH}}\nolimits ^ M_{k + r}(U, 1) \ar[r] & \mathop{\mathrm{CH}}\nolimits _{k + r}(X') \ar[r] & \mathop{\mathrm{CH}}\nolimits _{k + r}(X) \ar[r] & \mathop{\mathrm{CH}}\nolimits _{k + r}(U) \ar[r] & 0 \\ \mathop{\mathrm{CH}}\nolimits ^ M_ k(V, 1) \ar[r] \ar[u] & \mathop{\mathrm{CH}}\nolimits _ k(Y') \ar[r] \ar[u] & \mathop{\mathrm{CH}}\nolimits _ k(Y) \ar[r] \ar[u] & \mathop{\mathrm{CH}}\nolimits _ k(V) \ar[r] \ar[u] & 0 }$

Here we use the first higher Chow groups of $V$ and $U$ and the six term exact sequences constructed in Remark 42.27.3, as well as flat pullback for these higher chow groups and compatibility of flat pullback with these six term exact sequences. Since $U = \mathbf{A}^ r_ V$ the vertical map on the right is an isomorphism. The map $\mathop{\mathrm{CH}}\nolimits _ k(Y') \to \mathop{\mathrm{CH}}\nolimits _{k + r}(X')$ is bijective by induction on $d$. Hence to finish the argument is suffices to show that

$\mathop{\mathrm{CH}}\nolimits ^ M_ k(V, 1) \longrightarrow \mathop{\mathrm{CH}}\nolimits ^ M_{k + r}(U, 1)$

is surjective. Arguing as in the proof of Lemma 42.32.1 this reduces to Step 3 below.

Step 3. Let $F$ be a field. Then $\mathop{\mathrm{CH}}\nolimits ^ M_0(\mathbf{A}^1_ F, 1) = 0$. (In the proof of the lemma cited above we proved analogously that $\mathop{\mathrm{CH}}\nolimits _0(\mathbf{A}^1_ F) = 0$.) We have

$\mathop{\mathrm{CH}}\nolimits ^ M_0(\mathbf{A}^1_ F, 1) = \mathop{\mathrm{Coker}}\left( \partial : K^ M_2(F(T)) \longrightarrow \bigoplus \nolimits _{\mathfrak p \subset F[T]\text{ maximal}} \kappa (\mathfrak p)^*\right)$

The classical argument for the vanishing of the cokernel is to show by induction on the degree of $\kappa (\mathfrak p)/F$ that the summand corresponding to $\mathfrak p$ is in the image. If $\mathfrak p$ is generated by the irreducible monic polynomial $P(T) \in F[T]$ and if $u \in \kappa (x)^*$ is the residue class of some $Q(T) \in F[T]$ with $\deg (Q) < \deg (P)$ then one shows that $\partial (Q, P)$ produces the element $u$ at $\mathfrak p$ and perhaps some other units at primes dividing $Q$ which have lower degree. This finishes the sketch of the proof.

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