Proof. As $R$ is regular, it is a finite direct product of regular semi-local $k$-algebras essentially of finite type, each of which is an integral domain. We can therefore assume without loss of generality that $R$ is an integral domain.

As $R$ is essentially of finite type over $k$ and semi-local, we can find an integral domain $\bar {R}$, which is a $k$-algebra of finite type, and a finite set $\Sigma$ of primes in $\bar {R}$ such that $R = \bar {R}_{\Sigma }$, the localization at $\Sigma$. As $R$ is regular and the regular locus of ${{\rm Spec}}(\bar {R})$ is open, we can assume that $\bar {R}$ is regular.

We can now find a subfield $k' \subset k$ which is finitely generated over $\mathbb {F}_p$, a $k'$-algebra $\bar {A}$ of finite type such that $\bar {R} = \bar {A} \otimes _{k'} k$. As the prime ideals lying in the finite set $\Sigma$ are finitely generated, we can assume (after replacing $k'$ by a bigger finitely generated subfield of $k$) that there is a finite set $\Sigma '$ of primes in $\bar {A}$ such that $\Sigma = \{\mathfrak {p} \bar {R} \mid \mathfrak {p} \in \Sigma '\}$. By replacing the elements of $\Sigma '$ by the contractions of the elements in $\Sigma$, we can also assume that $\Sigma ' = \{\mathfrak {p} \cap \bar {A}| \mathfrak {p} \in \Sigma \}$.

As $\bar {R}$ is a regular $\mathbb {F}_p$-algebra and $\mathbb {F}_p$ is perfect, we see that $R$ is geometrically regular over $\mathbb {F}_p$. As the inclusion $\bar {A} \hookrightarrow \bar {R}$ of Noetherian $\mathbb {F}_p$-algebras is faithfully flat, the ring $\bar {A}$ is also geometrically regular over $\mathbb {F}_p$ (see e.g. [Stacks, Lemma 07NH]). In particular, it is regular.

Setting $R' = {\bar {A}}_{\Sigma '}$ and $R'_{k} = R' \otimes _{k'} k$, we have the canonical injection $R'_k \hookrightarrow \bar {R}_{\Sigma } = R$ which is a localization map. Writing $k$ as a direct limit of its finitely generated subfields, we can find a direct system of finitely generated subfields $\{k_i\}_{i \in I}$ such that $k' \subset k_i \subset k_{j} \subset k$ for $i \le j$, and $k = {\bigcup} _{i \in I} k_i$. In particular, $R'_k = {\varinjlim _i} R'_i$, where $R'_i:= R' \otimes _{k'} k_i$. As $R'_{j} = R' \otimes _{k'} k_{j} = (R' \otimes _{k'} k_i) \otimes _{k_i} k_{j} = R'_i \otimes _{k_i} k_{j}$ for $i \le j$, it follows that each transition map $\lambda _{ij} \colon R'_i \to R'_{j}$ is an injective faithfully flat map of Noetherian $\mathbb {F}_p$-algebras such that $R'_k =\varinjlim _i R'_i$. Furthermore, $R'_{k} = R'_i \otimes _{k_i} k$ and, hence, the inclusion $R'_i \hookrightarrow R'_k$ is also faithfully flat. In particular, each $R'_i$ is a regular integral domain.

As $R'$ is essentially of finite type over $k'$ and $k'$ is finitely generated over $\mathbb {F}_p$, it follows that $R'$ is an $\mathbb {F}_p$-algebra essentially of finite type. As each $k_i$ is finitely generated over $\mathbb {F}_p$, the same argument implies that each $R'_i$ is an regular $\mathbb {F}_p$-algebra essentially of finite type. As $\mathbb {F}_p$ is perfect, it follows that each $R'_i$ is smooth over $\mathbb {F}_p$. We let $\Sigma _i = \{\mathfrak {p} \cap R'_i| \mathfrak {p} \in \Sigma \}$ under the inclusion $R'_i \hookrightarrow R$. It is clear that for each pair of elements $i \le j$ in $I$, the transition map $\lambda _{ij}: R'_i \hookrightarrow R'_{j}$ induces an inclusion $\lambda _{ij} \colon (R'_i)_{\Sigma _i} \hookrightarrow (R'_j)_{\Sigma _j}$ such that $\Sigma _i = \{\mathfrak {p} \cap (R'_i)_{\Sigma _i}| \mathfrak {p} \in \Sigma _j\}$. We let $R_i = (R'_i)_{\Sigma _i}$ for $i \in I$.

As each $\lambda _{ij}$ is flat and every closed point of ${{\rm Spec}}(R_i)$ is in the image of the map $\lambda ^{*}_{ij} \colon {{\rm Spec}}(R'_j) \to {{\rm Spec}}(R_i)$, it follows that $\lambda _{ij}$ is faithfully flat (see e.g. [Stacks, Lemma 00HQ] or [Reference MatsumuraMat86, Theorem 7.2, p. 47]). The same reasoning implies that each inclusion $R_i \hookrightarrow R$ is faithfully flat.

Finally, we note that every element of $R'_k$ which becomes invertible in $R$ lies in the image of $R'_i \hookrightarrow R$ for some $i \in I$. Such an element must become invertible in $R_i$. It follows that $R =\varinjlim _i R_i$. This finishes the proof.

Proof. Let $\bar {R}$ be the regular finite type $k$-algebra and $\Sigma$ a finite subset of ${{\rm Spec}}(\bar {R})$ as in the proof of Lemma 2.3 such that $R = \bar {R}_{\Sigma }$. We can similarly find a regular finite type $k$-algebra $\bar {S}$ and a finite subset $\Phi \subset {{\rm Spec}}(S)$ such that $S = \bar {S}_\Phi$. As $\bar {R}$ is a finite type $k$-algebra, we can extend the map $f \colon R \to S$ to a $k$-algebra morphism $f \colon \bar {R} \to \bar {S}$ (after possibly replacing $\bar {S}$ by a localization which is still finite type over $k$).

Next, we saw in the proof of Lemma 2.3 that we can find a subfield $k' \subset k$ which is finitely generated over $\mathbb {F}_p$, a finite-type $k'$-algebra $\bar {A}$ (respectively, $\bar {B}$), and a finite subset $\Sigma ' \subset {{\rm Spec}}(\bar {A})$ (respectively, $\Phi ' \subset {{\rm Spec}}(\bar {B})$) such that $\bar {R} = \bar {A} \otimes _{k'} k$ (respectively, $\bar {S} = \bar {B} \otimes _{k'} k$) and $\Sigma$ (respectively, $\Phi$) is the extension of $\Sigma '$ (respectively, $\Phi '$). Furthermore, $R = R' \otimes _{k'} k$, where $R' = {\bar {A}}_{\Sigma '}$. Similarly, $S = S' \otimes _{k'} k$, where $S' = \bar {B}_{\Phi '}$.

As $\bar {A}$ is a finite-type $k'$-algebra, we can find a field $k''$ such that it is finitely generated over $\mathbb {F}_p$ with $k' \subset k'' \subset k$ and a $k''$-algebra morphism $f \colon \bar {A} \otimes _{k'} k'' \to \bar {B} \otimes _{k'} k''$ which extends $f \colon \bar {R} = \bar {A} \otimes _{k'} k \to \bar {B} \otimes _{k'} k = \bar {S}$. In other words, we can choose our finitely generated subfield $k' \subset k$ over $\mathbb {F}_p$ such that $f \colon \bar {R} \to \bar {S}$ is induced by a morphism of finite type $k'$-algebras $f \colon \bar {A} \to \bar {B}$.

As, after base change to $k$, the map $f \colon \bar {A} \to \bar {B}$ induces the map $R'_k = {\bar {A}}_{\Sigma '} \otimes _{k'} k \to {\bar {B}}_{\Phi '} \otimes _{k'} k = S'_k$, it follows that $f$ must induce a map of localizations $f \colon R' = {\bar {A}}_{\Sigma '} \to {\bar {B}}_{\Phi '} = S'$. We have therefore shown that there is a finitely generated subfield $k' \subset k$ over $\mathbb {F}_p$, a morphism between essentially of finite-type regular semi-local $k'$-algebras $f \colon R' \to S'$ which induces our original map $f \colon R = (R'_k)_{\Sigma } \to (S'_k)_{\Phi } = S$.

We now choose a direct system of finitely generated subfields $\{k_i\}_{i \in I}$ such that $k' \subset k_i \subset k_{j} \subset k$ for every $i \le j$, and $k = {\bigcup} _{i \in I}\, k_i$. We let

\[ R'_i = R \otimes_{k'} k_i, \quad S'_i = S' \otimes_{k'} k_i,\! \quad \Sigma_i = \Sigma \cap R'_i,\! \quad \Phi_i = \Phi \cap S'_i,\! \quad R_i = (R'_i)_{\Sigma_i},\! \quad \mbox{and} \!\quad S_i = (S'_i)_{\Phi_i}. \]

The rings $R$ and $S$ are semi-local with the sets of maximal ideals $\Sigma$ and $\Phi$, respectively. It is easy to check using this that the map $f_i \colon R'_i \to S'_i$ induced by $f \colon R' \to S'$ descends to a map between the localizations $f_i \colon (R'_i)_{\Sigma _i} \to (S'_i)_{\Phi _i}$. Equivalently, a map $f_i \colon R_i \to S_i$. The rest of the proof now is a repetition of the proof of Lemma 2.3.

Proof. It follows, e.g., from [Reference RüllingRül07a, Proposition 1.16 and Lemma 1.17].

Proof. By [Reference HesselholtHes15, Definitions 3.1, 3.6 and p. 186], we have $d[a] \wedge d [a] = d\log [-1] \wedge F_2 ( d [a])$ for each $a \in R$. If ${\rm char} (k) = 2$, then $-1 = 1$ in $k$ so that $d\log [-1] = d\log [1] = 0$. Hence, $d[a] \wedge d [a] = d\log [-1] \wedge F_2 (d [a]) = 0$. If ${\rm char} (k) \not = 2$, then, by the anti-commutativity, we have $d [a] \wedge d [a] = - d [a] \wedge d[a]$ so that $2 d [a] \wedge d [a] = 0$. As $2 \in k^{\times }$ so that $2 \in \mathbb {W}_m (k)^{\times }$, we deduce $d [a] \wedge d [a] = 0$.

Proof. Let $K$ denote the total ring of quotients of $R$. This is a product of fields because $R$ is regular and, hence, reduced. Using Néron–Popescu (Theorem 2.2 in § 2.6), we can write $R = {\varinjlim }_{\lambda \in I} R_\lambda$ such that each $R_\lambda$ is a semi-local ring which is essentially of finite type and smooth over the prime field of $R$. As taking the total rings of quotients commutes with a filtering inductive limit of rings, it follows that $K = {\varinjlim }_{\lambda \in I} K_\lambda$, where $K_\lambda$ is the total ring of quotients of $R_\lambda$.

We now have a commutative diagram of canonical maps

(2.4)

in which the horizontal arrows are isomorphisms by [Reference RüllingRül07a, Proposition 1.16]. It suffices therefore to prove the proposition when $k$ is perfect and $R$ is a semi-local ring which is essentially of finite type and smooth over $k$. In this case, $R$ is a finite product of integral domains. We can therefore assume further that $R$ is an integral domain.

We prove this case in two steps.

Step 1. Let $m=1$. We show that $\Omega _{R/\mathbb {Z}} ^{n} \to \Omega _{K/ \mathbb {Z}} ^{n}$ is injective.

Claim We claim that $\Omega ^{i} _{R/ \mathbb {Z}}$ is a free $R$-module, possibly of infinite rank.

Proof. This is an immediate consequence of [Reference Krishna and ParkKP17a, Theorem 3.12].

Proof. For the $(\Leftarrow )$ direction, one sees that the existence of a smooth atlas $(X,\Sigma )$ for $V$ and a cycle $\bar {Z} \in {{\operatorname {Tz}}}^{n}_\Sigma (X, n; m)$ satisfying properties (i)–(iv) imply that $Z \in {{\operatorname {Tz}}}^{n} _\Sigma (V, n; m)$ is an ${{\rm sfs}}$-cycle over $V$. Thus, we need to prove the converse $(\Rightarrow )$.

Suppose that $Z \in {{\operatorname {Tz}}}^{n} _{\Sigma } (V, n ; m)$ is an irreducible sfs-cycle. As it is an fs-cycle by definition, applying [Reference Krishna and ParkKP20, Lemma 2.9, Proposition 2.13, Lemma 2.21], we see that there is a smooth affine atlas $(X, \Sigma )$ for $V$ such that the closure $\bar {Z}$ of $Z$ in $X \times B_n$ is an irreducible admissible cycle in ${{\operatorname {Tz}}}^{n} _\Sigma (X, n;m)$ which satisfies properties (i) and (ii).

As $\bar {Z} \to X$ is fs over $X$, there is a sequence of finite maps $\bar {Z} = {\bar {Z}}^{(n)} \to \cdots \to {\bar {Z}}^{(1)} \to X = {{\rm Spec}}(A)$ such that each ${\bar {Z}}^{(j)}$ is fs over $X$. Furthermore, each $\bar {Z}^{(j)} _V$ is smooth over $k$ because $Z$ is an ${{\rm sfs}}$-cycle.

We now want to show that there is a smooth affine open subatlas $(U,\Sigma )$ of $(X,\Sigma )$ for $V$ such that the restrictions $\bar {Z}^{(j)} _U$ to $U$ $1 \leq j \leq n-1$ are all smooth over $k$. To prove it, set $A^{j} = k [\bar {Z} ^{(j)}]$. As each $\bar {Z} ^{(j)}_V$ is smooth over $k$ and finite over $V$, we see that $\Omega _{A^{j}/k} ^{1}$ is a finite $A$-module such that $\Omega _{S_\Sigma ^{-1} A^{j}/k} ^{1}$ is a free $R$-module, where $R = S_\Sigma ^{-1} A$ for the multiplicative subset $S_\Sigma \subset A$ corresponding to the finite set of closed points $\Sigma$. Hence, there is an affine open neighborhood $U$ of $\Sigma$ in $X$ such that each $\Omega ^{1} _{A^{j}/k} |_U$ is a free $k[U]$-module. Replacing the given atlas $(X, \Sigma )$ by the new $(U,\Sigma )$, we may thus assume that each $\bar {Z}^{(j)}$ is smooth over $k$. Hence, we have proven property (iii).

To prove property (iv), we observe that if we replace $A$ by its semi-local ring $R$ via localization, then we obtain a sequence of finite extensions of smooth semi-local rings. Note that such rings are unique factorization domains (UFDs) by Auslander–Buchsbaum and $R[a] = R[t]/ I_1$ for the prime ideal $I_1= I ({Z}^{(1)})$. As $\dim (R)= \dim (R[a]) = \dim (R[t]) -1$, we have ${\rm ht} (I_1) = 1$. However, $R[t]$ is a UFD so that $I_1$ must be principal by [Reference MatsumuraMat86, Theorem 20.1, p. 161]. Thus, if $P(t) \in R[t]$ is a monic irreducible polynomial of $a$, then we have $I_1 = (P(t))$. Similarly, we have $R[a,b_1] = R[a][y_1]/I_2$ for the prime ideal $I_2 = I({Z} ^{(2)})$, and because $R[a]$ is a UFD, so is $R[a][y_1]$. Hence, we obtain $I_2 = (Q_1 (y_1))$ in the same way.

Continuing as above, we get the irreducible monic polynomials $P(t) \in R[t]$ and $Q_j(y_j) \in R[a, b_1, \ldots , b_{j-1}][y_j]$ for which property (iv) holds over $R$. Choose lifts of these polynomials over $A$, and then there is a localization $A' = A[f^{-1}]$ for some $f \in A$ with the inclusions $A \hookrightarrow A' \hookrightarrow R$ such that the property (iv) holds over $A'$. Replacing $(X,\Sigma )$ by $({{\rm Spec}} (A'), \Sigma )$, we obtain a new atlas for $V$ for which all of properties (i)–(iv) hold.

Proof. We assume first that $k$ is perfect. Let $V= {{\rm Spec}} (R)$. If $p \neq 2$, then the lemma is [Reference Krishna and ParkKP16, Theorem 1.2]. Thus, we assume that $p =2$. In this case, if one closely follows the arguments of [Reference Krishna and ParkKP15] and [Reference Krishna and ParkKP16], then one will notice that there is exactly one place where we used the assumption $p \neq 2$, namely, to show that $\delta ^{2} = 0$ for the cycle-theoretic differential operator $\delta$ on ${{\operatorname {TCH}}}^{\bullet }(R;m) = {{\operatorname {TCH}}}^{\bullet }(V;m)$ (see [Reference Krishna and ParkKP16, Definition 6.2]). However, it turns out that this vanishing actually follows from the other properties of ${{\operatorname {TCH}}}^{\bullet }(V;m)$ proven in [Reference Krishna and ParkKP15] and [Reference Krishna and ParkKP16] without using any assumption on $p$. We explain this in the following.

Let $\mathcal {R} \colon {{\operatorname {TCH}}}^{\bullet }(V;m+1) \to {{\operatorname {TCH}}}^{\bullet }(V;m)$ denote the obvious restriction map. Let $F_r \colon {{\operatorname {TCH}}}^{\bullet }(V;rm+r-1) \to {{\operatorname {TCH}}}^{\bullet }(V;m)$ denote the Frobenius map given by the push-forward via the map $\phi _r \colon \mathbb {A}^{1} \to \mathbb {A}^{1}$, where $\phi _r(t) = t^{r}$ (see [Reference Krishna and ParkKP16, § 7.1]). Let $V_r \colon {{\operatorname {TCH}}}^{\bullet }(V;m) \to {{\operatorname {TCH}}}^{\bullet }(V;rm+r-1)$ be the Verschiebung map given by the flat pull-back via $\phi _r$. It was shown in the proofs of [Reference Krishna and ParkKP15, Theorem 5.13] and [Reference Krishna and ParkKP16, Theorem 7.1] that $F_r, V_r$ and $\delta$ satisfy the identity

(4.2)\begin{equation} F_r \delta V_r = \delta \quad \forall \, r \ge 1 \end{equation}

on the additive higher Chow groups of $V$.

As part of the proof of this identity, it was also shown (as the first claim) in the proof of [Reference Krishna and ParkKP15, Theorem 5.13, Part 2, p. 45] that $F_r$ and $\delta$ satisfy the identity

(4.3)\begin{equation} rF_r \delta = \delta F_r \quad \forall \, r \ge 1 \end{equation}

on the additive higher Chow groups of $V$. Furthermore, the proofs of the above two identities do not use any condition on $k$. (Recall from [Reference HesselholtHes15, Lemma 4.3] that (4.3) is one of the properties of a restricted Witt complex.) We now show that these two identities are enough to conclude that $\delta ^{2} = 0$.

Suppose, in general, that $m \ge 0$ is an integer and let $\alpha \in {{\operatorname {TCH}}}^{q} (V, n;m)$ be a cycle class. Then the normalization theorem [Reference Krishna and ParkKP16, Theorem 3.2] says that $\alpha$ is represented by a cycle (also denoted by $\alpha$) such that $\partial _i ^{\epsilon } (\alpha ) = 0$ for all $1 \leq i \leq n-1$ and $\epsilon \in \{ 0, \infty \}$. It follows therefore from [Reference Krishna and ParkKP16, Lemma 6.5] that $2 \delta ^{2} (\alpha ) = 0$. Combining this with (4.3), we get $\delta F_2 \delta (\alpha ) = 2 F_2 \delta ^{2}(\alpha ) = F_2(2 \delta ^{2}(\alpha )) = 0$. As $\alpha$ and $m \ge 0$ were arbitrary, we conclude that $\delta F_2 \delta = 0$ on ${{\operatorname {TCH}}}^{\bullet }(V;m)$ for all $m \ge 0$. In particular, $\delta F_2 \delta V_2 = 0$ on ${{\operatorname {TCH}}}^{\bullet }(V;m)$ for all $m \ge 0$.

We now use (4.2) to conclude $\delta ^{2} = \delta (F_2 \delta V_2) = \delta F_2 \delta V_2 = 0$. This finishes the proof of the lemma when $k$ is perfect.

To prove the lemma when $k$ is imperfect, the only thing we need to recall for the reader is that in the construction of the Witt-complex structure on additive higher Chow groups in [Reference Krishna and ParkKP15] and [Reference Krishna and ParkKP16], the perfectness of the ground field was required exactly at one place, namely, to construct the product structure on the additive Chow groups of smooth schemes. More specifically, this was needed only in the proof of [Reference Krishna and ParkKP15, Lemma 2.11]. However, as pointed out by the referee, the proof of the latter result actually goes through without the perfectness assumption. The improved version is Lemma 4.2 below. The rest of the proof of the Witt-complex structure is identical to the perfect base field case.

Proof. Let $W \subset V_1 \times V_2$ be an irreducible component. It is enough to show that $W$ satisfies the modulus $(m_1, m_2)$ condition. Let $\bar {V}_1 \subset X \times \bar {B}_{n_1}$ and $\bar {V}_2 \subset Y \times \bar {B}_{n_2}$ be the Zariski closures of $V_1$ and $V_2$, respectively. Then $\bar {W} \subset X \times Y \times \bar {B}_{2, n_1 + n_2}$ is an irreducible component of $\bar {V}_1 \times \bar {V}_2$. Let $\bar {V} = (\bar {V}_1 ^{N} \times \bar {V}_2 ^{N})_{{\rm red}}$.

We now note that $\bar {V}_1 \times \bar {V}_2$ may be neither reduced nor irreducible. Nonetheless, it is equidimensional and the projection $\bar {W} \to \bar {V}_i$ is dominant for each $i$. It follows that $\bar {V}^{N}$ is equidimensional with $\bar {W}^{N}$ one of its irreducible components and the projection $\bar {W}^{N} \to \bar {V}^{N}_i$ is dominant for each $i$. This gives rise to a commutative diagram

(4.4)

where $\iota _1 = (\nu _{\bar {V}_1} \times {\rm Id})$ and $\iota _2 = ({\rm Id} \times \nu _{\bar {V}_2})$. Let $(t_1, t_2, y_1, \ldots , y_{n-1}) \in \mathbb {A}^{2} \times ( \mathbb {P} ^{1} )^{n-1}$ be the coordinates. Then, for $D^{1}: = \sum _{i=1 } ^{n_1 -1} \{ y_i = 1 \} - (m_1 +1) \{ t_1 = 0 \}$, $D^{2}:= \sum _{i={n_1}} ^{ n-1} \{ y_i = 1 \} - (m_2 +1) \{ t_2 = 0 \}$, and $n-1 = (n_1 -1) + (n_2 -1)$, we have

(4.5)\begin{equation} \iota^{*}_1 (D^{1}) \geq 0 \quad \mbox{and} \quad \iota^{*}_2 (D^{2}) \geq 0. \end{equation}

We should note here that $\bar {V}^{N}_1 \times Y \times \bar {B}_{n_2}$ and $X \times \bar {B}_{n_1} \times \bar {V}^{N}_2$ are normal (this is ensured by the smoothness of $X$ and $Y$).

As $\bar {W}^{N} \to \bar {V}^{N}_i$ is dominant, we see that $v^{*}_1 \circ \iota ^{*}_1 (\sum _{i=1 } ^{n_1 -1} \{ y_i = 1 \})$ and $v^{*}_1 \circ \iota ^{*}_1((m_1 +1) \{ t_1 = 0 \})$ are effective Cartier divisors on $\bar {W}^{N}$. It follows from (4.5) that $v^{*}_1 \circ \iota _1^{*} (D^{1}) \geq 0$. Similarly, we have $v^{*}_2 \circ \iota ^{*}_2(D^{2}) \ge 0$. Combining the two, we get $\nu ^{*}_{\bar {W}}(D^{1}+ D^{2}) \ge 0$, which is the modulus $(m_1, m_2)$ condition for $W$.

Proof. The lemma is equivalent to proving the following:

1. (1) $f^{*}$ commutes with products;

2. (2) $f^{*}$ commutes with differentials;

3. (3) $f^{*}$ commutes with $\mathcal {R}$, $F_r$, and $V_r$; and

4. (4) $f^{*}$ commutes with $\lambda _R$.

Part (3) is shown in [Reference Krishna and ParkKP16, Theorem 7.1] while part (4) is evident from (4.1). We prove parts (1) and (2).

Let $V = {{\rm Spec}}(R)$ and $V' = {{\rm Spec}}(R')$. To prove part (1), it suffices to prove it for irreducible cycles. By [Reference Krishna and ParkKP12a, Proof of Theorem 7.1], there exists a finite set $\mathcal {W}$ of locally closed subsets of $V$ such that the map $f^{*}\colon {{\operatorname {Tz}}}^{q}_{\mathcal {W}}(V, \bullet ;m) \to {{\operatorname {Tz}}}^{q}(V', \bullet ;m)$ given by $f^{*}(Z) = [f^{-1}(Z)]$, is well-defined. Here, the group on the left is defined as in Definition 3.2. Choose an irreducible cycle $Z \in {{\operatorname {Tz}}}^{q}_{\mathcal {W}}(V, \bullet ;m)$. We claim that there is a finite set $\mathcal {C}$ of locally closed subsets of $V$ such that the following hold.

1. (i) $f^{*}\colon {{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m) \to {{\operatorname {Tz}}}^{q}(V', \bullet ;m)$ is defined;

2. (ii) $Z \boxtimes Z' \in {{\operatorname {Tz}}}^{q}_{\{\Delta _V\}}(V \times V, \bullet ;m)$ for all $Z' \in {{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m)$;

3. (iii) $f^{*}(Z) \boxtimes f^{*}(Z') \in {{\operatorname {Tz}}}^{q}_{\{\Delta _{X'}\}}(V' \times V', \bullet ;m)$ for all $Z' \in {{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m)$.

Let $Z_f$ be the (finite) collection $\{Z_i \cap (V' \times \mathbb {A}^{1} \times F)\}$, where $Z_i$ is an irreducible component of $f^{*}(Z)$ and $F \subset \square ^{n-1}$ is a face. Under the moving lemma [Reference Krishna and ParkKP16, Theorem 4.10], by the argument of [Reference Krishna and ParkKP17a, Lemmas 3.5 and 3.10], there exists a finite collection $\mathcal {W}'$ of locally closed subsets of $\{V \times \mathbb {A}^{1} \times \square ^{n_i}\}$ such that for ${{\operatorname {Tz}}}_{\mathcal {W}'} ^{q} (V, \bullet ;m)$ in the sense of [Reference Krishna and ParkKP12b, Definition 5.3], we have:

1. (i) $f^{*}\colon {{\operatorname {Tz}}}^{q}_{\mathcal {W}'}(V, \bullet ;m) \to {{\operatorname {Tz}}}^{q}_{Z_f}(V', \bullet ;m)$ is well-defined;

2. (ii) $Z \boxtimes Z' \in {{\operatorname {Tz}}}^{q}_{\{\Delta _X\}}(V \times V, \bullet ;m)$ for all $Z' \in {{\operatorname {Tz}}}^{q}_{\mathcal {W}'}(V, \bullet ;m)$;

3. (iii) $Z_i \boxtimes f^{*}(Z') \in {{\operatorname {Tz}}}^{q}_{\{\Delta _{V'}\}}(V' \times V', \bullet ;m)$ for all $Z' \in {{\operatorname {Tz}}}^{q}_{\mathcal {W}'}(V, \bullet ;m)$ and all irreducible components $Z_i$ of $f^{*}(Z)$.

Furthermore, it follows also by the moving lemma [Reference Krishna and ParkKP16, Theorem 4.10] and the argument of [Reference Krishna and ParkKP17a, Lemma 3.4] that there exists a finite collection $\mathcal {C}'$ of locally closed subsets of $V$ such that ${{\operatorname {Tz}}}^{q}_{\mathcal {W}'}(V, \bullet ;m) = {{\operatorname {Tz}}}^{q}_{\mathcal {C}'}(V, \bullet ;m)$. Setting $\mathcal {C} = \mathcal {W} \cup \mathcal {C}'$, we get the proof of the claim.

We now consider the commutative diagram

(4.6)

where $\mu \colon \mathbb {A} ^{1} \times \mathbb {A}^{1} \to \mathbb {A}^{1}$ is $(t_1, t_2) \mapsto t_1t_2$.

If we choose any irreducible cycle $Z' \in {{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m)$, then it follows from the above claim that $\Delta ^{*}_V(Z \boxtimes Z')$ and $\Delta ^{*}_{V'} \circ (f\times f)^{*}(Z \boxtimes Z')$ are admissible cycles. Moreover, by the commutativity of the right square in (4.6), we have $\Delta ^{*}_{V'} \circ (f\times f)^{*}(Z \boxtimes Z')= f^{*}(\Delta ^{*}_V(Z \boxtimes Z'))$. In particular, $f^{*} \circ \Delta ^{*}_V(Z \boxtimes Z')$ is an admissible cycle.

Now by [Reference Krishna and ParkKP16, Corollary 5.11], the cycles $\mu _* \circ \Delta ^{*}_{V'} \circ (f\times f)^{*}(Z \boxtimes Z')$ and $\mu _* \circ \Delta ^{*}_V \circ (f\times f)^{*}(Z \boxtimes Z')$ are admissible. As the left square in (4.6) is transverse, we obtain $f^{*}(Z \cdot Z') = f^{*} \circ \mu _* \circ \Delta ^{*}_V(Z \boxtimes Z') = \mu _* \circ \Delta ^{*}_V \circ (f\times f)^{*}(Z \boxtimes Z') = f^{*}(Z) \cdot f^{*}(Z').$

Finally, the inclusions ${{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m) \hookrightarrow {{\operatorname {Tz}}}^{q}(V, \bullet ;m)$ and ${{\operatorname {Tz}}}^{q}_{\mathcal {W}}(V, \bullet ;m) \hookrightarrow {{\operatorname {Tz}}}^{q}(V, \bullet ;m)$ are quasi-isomorphisms by [Reference Krishna and ParkKP16, Theorem 4.10], and this shows part (1).

To prove part (2), recall from [Reference Krishna and ParkKP16, Definition 6.2] that the differential $\delta =\delta _V \colon {{\operatorname {Tz}}}^{q}(V, n;m) \to {{\operatorname {Tz}}}^{q+1}(V,n+1;m)$ is defined as the push-forward with respect to the map

(4.7)\begin{equation} \begin{array}{c} \delta_V \colon V \times \mathbb{G}_m \times \square^{n-1} \to V \times \mathbb{A}^{1} \times \square^{n};\\ \delta (x, t, y_1, \ldots, y_{n-1}) = (x, t, t^{{-}1}, y_1, \ldots, y_{n-1}) \quad \mbox{for} \ t \neq 0, 1. \end{array} \end{equation}

Here, $\delta _{V*}(Z)$ is an irreducible admissible cycle if $Z$ is so, by [Reference Krishna and ParkKP16, Lemma 6.3, Proposition 6.4].

As the square

is transverse, it follows that $\delta _{V' *} \circ f^{*}(Z) = f^{*} \circ \delta _{V *}(Z)$ for every irreducible cycle $Z \in {{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m)$. We conclude again that $f^{*} \circ \delta = \delta \circ f^{*}$ from the quasi-isomorphism ${{\operatorname {Tz}}}^{q}_{\mathcal {C}}(V, \bullet ;m) \hookrightarrow {{\operatorname {Tz}}}^{q}(V, \bullet ;m)$. The proof of the lemma is now complete.

Proof. As the statement of the lemma is local on $X \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k$, we can replace $\overline {\square }^{n-1}_k \simeq \mathbb {P}^{n-1}_k$ by $\mathbb {A}^{n-1}_k$. The problem then reduces to showing that the map $R[t, y_1, \ldots , y_{n-1}] \to R'[t, y_1, \ldots , y_{n-1}]$ is faithfully flat. This is obvious because $R \to R'$ is faithfully flat.

Proof. The proofs for the higher Chow groups and additive higher Chow groups are identical. Thus, we prove the latter case only. As the homology functor commutes with the direct limit, it suffices to show that the lemma holds on the level of cycle complexes. We set $V_i = {{\rm Spec}}(R_i)$ and $V = {{\rm Spec}}(R)$. Let $\lambda '_i: R_i \to R$ be the natural map. This is also faithfully flat; this follows from the fact that a direct limit of flat modules is flat, and an $R_i$-module $M$ is faithfully flat if and only if it is flat and $\mathfrak {m} M \neq 0$ for every nonzero maximal ideal $\mathfrak {m} \subset R_i$ (see [Reference MatsumuraMat86, Theorem 7.2, p. 47]).

The projections ${{\rm Spec}}(K) \to {{\rm Spec}}(k)$ and $V \to V_j \to V_i$ between $k$-schemes induce the projection maps $V \times _K \mathbb {A}^{1}_K \times _K \overline {\square }^{n-1}_K \to V_i \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k$. This morphism is faithfully flat. It induces the pull-back maps between the cycle groups ${{\operatorname {Tz}}}^{q}(V_i, n;m) \xrightarrow {\lambda ^{*}_{ij}} {{\operatorname {Tz}}}^{q}(V_j, n;m) \xrightarrow {\lambda ^{\prime *}_{j}} {{\operatorname {Tz}}}^{q} (V, n;m)$.

The injectivity of the map ${\varinjlim _i} {{\operatorname {Tz}}}^{q}(V_i, \bullet ;m) \to {{\operatorname {Tz}}}^{q}(V, \bullet ;m)$ is obvious. To show its surjectivity, let $Z \in {{\operatorname {Tz}}}^{q}(V, n;m)$ be an irreducible admissible cycle and let $\bar {Z} \subset V \times _K \mathbb {A}^{1}_K \times _K \overline {\square }^{n-1}_K$ be its Zariski closure and $\nu _Z: \bar {Z}^{N} \to V \times _K \mathbb {A}^{1}_K \times \overline {\square }^{n-1}_K$ the normalization map. As each of $f_i := {{\rm Spec}}(\lambda _i) : V_{i+1} \to V_i$ and $f'_i := {{\rm Spec}}(\lambda '_i) : V \to V_i$ is faithfully flat, by Lemma 4.5, the product maps $\widetilde {f_i}: V_{i+1} \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k \to V_i \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k$ and $\widetilde {f'_i}: V \times _K \mathbb {A}^{1}_K \times _K \overline {\square }^{n-1}_K \to V_i \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k$ are faithfully flat maps of Noetherian $k$-schemes.

As $V = \varprojlim _i V_i$, it follows from the above faithfully flatness that there exists $i \gg 0$ and an irreducible cycle $\bar {Z}_i \hookrightarrow V_i \times _k \mathbb {A}^{1}_k \times _k \overline {\square }^{n-1}_k$ such that $(\widetilde {f'_i})^{*}(\bar {Z}_i) = \bar {Z}$ and $Z_i := \bar {Z}_i \cap (V_i \times _k \mathbb {A}^{1}_k \times _k {\square }^{n-1}_k) \in z^{q}(V_i \times _k \mathbb {A}^{1}_k, n-1)$.

As the right square in the following commutative diagram

(4.8)

is Cartesian, we must have $Z = (\widetilde {f'_i})^{*}({Z_i})$.

As $\widetilde {f'_i}$ is faithfully flat, it follows that $\bar {Z} = (\widetilde {f'_i})^{*}(\bar {Z}_i) \to \bar {Z}_i$ is also faithfully flat. In particular, it is surjective. We conclude that the induced map $\bar {Z}^{N} \to \bar {Z}_i^{N}$ on the normalizations is also surjective. As $\bar {Z}$ satisfies the modulus condition, we can now apply Lemma 4.4 to conclude that $Z_i$ also satisfies the modulus condition and, hence, it is in ${{\operatorname {Tz}}}^{q}(V_i, n;m)$. This finishes the proof.

Proof. If $k$ is finite, the theorem follows from Lemmas 4.1 and 4.3. Thus, we may now assume that $k$ is infinite and $p > 1$.

Let $\{R_i\}_{i \in I}$ be the direct system of subrings in $R$ as in Lemma 2.3. As each transition map $\lambda _{ij} \colon R_i \to R_{j}$ is flat, we have the pull-back maps of projective systems $\lambda ^{*}_{ij} \colon ({{\operatorname {TCH}}}^{\bullet }(R_i; m))_{m \in \mathbb {N}} \to ({{\operatorname {TCH}}}^{\bullet }(R_{j}; m))_{m \in \mathbb {N}}$.

As each $R_i$ is a smooth $\mathbb {F}_p$-algebra essentially of finite type, it follows from Lemma 4.1 that $({{\operatorname {TCH}}}^{\bullet }(R_i; m))_{m \in \mathbb {N}}$ is a restricted Witt complex over $R_i$. Furthermore, it follows from Lemma 4.3 that each $\lambda ^{*}_{ij}$ is a morphism of restricted Witt complexes over $R_i$. Thus, we conclude from Lemma 2.6 that $(\varinjlim _i {{\operatorname {TCH}}}^{\bullet }(R_i; m))_{m \in \mathbb {N}}$ forms a restricted Witt complex over $R$. The first part of the theorem now follows Lemma 4.6. The second part follows directly by combining Lemmas 2.5, 4.3, and 4.6.

Proof. Note that $f$ must be flat in our case (see [Reference HartshorneHar77, Exercise III-10.9, p. 276] or [Reference GrothendieckEGAIV, Proposition (6.1.5), p. 136]). The first part is [Reference Krishna and ParkKP16, Proposition 7.3] and the second part is [Reference Krishna and ParkKP15, Theorem 3.10] (whose proof does not require projectivity of the underlying schemes if the morphisms between them is finite and flat).

Proof. The left vertical arrow is a morphism of restricted Witt complexes over $R$ by Theorem 4.7. The maps $\tau ^{R}_\bullet$ and $\tau ^{R'}_\bullet$ are morphisms of restricted Witt complexes over $R$. It follows that $f^{*} \circ \tau ^{R}_\bullet$ and $\tau ^{R'}_\bullet \circ f^{*}$ are both morphisms of restricted Witt complexes over $R$. Using (4.1), the diagram (5.2) is easily seen to commute when $n = 1$. As $\mathbb {W}_m\Omega ^{\bullet }_R$ is the universal restricted Witt complex over $R$, we must have $f^{*} \circ \tau ^{R}_\bullet = \tau ^{R'}_\bullet \circ f^{*}$ on $\mathbb {W}_m\Omega ^{\bullet }_R$.

Proof. We take $R'$ to be the total ring of quotients of $R$ and apply Propositions 2.8 and 5.1. As $R'$ is a product of fields, by the commutative diagram (5.2) we are reduced to proving the lemma when $R$ is a field. However, this case follows from [Reference RüllingRül07a, Theorem 1] (when $p \ne 2$) and Theorem A.1 of the appendix (when $p = 2$).

Proof. Its proof is similar to that of [Reference RüllingRül07a, Lemma 2.20], for instance. Let $\omega = (\omega _n)_{n \in T} \in \mathbb {W}_T (S)$. Suppose $\omega \not = 0$, as otherwise there is nothing to prove. Define an operator $\varphi$ as follows: first choose $s_0 = \min \{ s \in T \mid \omega _s \not = 0 \}$. This minimum exists because $\omega \not = 0$. Here, $\omega _{s_0} \in S$ so that there exists a unique expression $\omega _{s_0} = \sum _{i=1} ^{e} c_{s_0, i} \cdot x_i$ in $S$ for some $c_{s_0, i} \in R$. Define $\varphi (\omega ):= \omega - V_{s_0} (\sum _{i=1} ^{e} [ c_{s_0, i}]_{T/ s_0} \cdot [ x_i ] _{T/ s_0})$. We have either $\varphi (\omega ) = 0$ or $\varphi (\omega ) \not = 0$. In the former case, the argument stops, whereas in the latter case, there exists $s_1: = \min \{ s \in T \mid \varphi (\omega )_s \not = 0 \}$. By construction, we have $s_1 > s_0$. We repeat this procedure. As $|T|< \infty$, there exists $N \geq 1$ such that eventually $\varphi ^{N} (\omega ) = 0$.

Proof. Using Proposition 4.10 and Lemma 5.7, it remains to check that $\tau ^{R}_{1,m} ({{\rm Tr}}_{S/R} ([x])) = f_* ([ \Gamma _{(1-xt)}])$ for all $x \in S$, where $\Gamma _{(1-xt)}$ is the cycle in ${{\operatorname {TCH}}}^{1} (S, 1;m)$ corresponding to the ideal $(1-xt) \subset S[t]$. As $[x] \in \mathbb {W}_m (S)$ corresponds to $1-xt \in (1+ t S[[t]])^{\times } / (1 + t^{m+1} S[[t]])^{\times }$, by the definition of ${{\rm Tr}}_{S/R}$, we have ${{\rm Tr}}_{S/R} ([x]) = {\rm N}_{S/R} (1 - xt)$. On the other hand, for a polynomial representative $g(t) \in (1 + t R[[t]])^{\times } / (1 + t^{m+1} R[[t]])^{\times }$, we have $\tau _{1,m} ^{R} ( g(t)) = [\Gamma _{(g(t))}]$, by definition. Hence, $\tau _{1,m} ^{R} ({{\rm Tr}}_{S/R} ([x])) = \tau _{1,m} ^{R} ({\rm N}_{R/S} (1-xt)) = [\Gamma _{({\rm N}_{S/R} (1-xt))}]$.