hcju/stacksqa huggingface dataset:
Let $Z \subset Y$ be an integral closed subscheme of $\delta$-dimension $n - 1$. We want to show that the coefficient of $[Z]$ in $p_*\text{div}(f)$ and $\text{div}(g)$ are equal. We may apply Lemma \ref{lemma-finite-in-codimension-one} to the morphism $p : X \to Y$ and the generic point $\xi \in Z$. Hence we may replace $Y$ by an affine open neighbourhood of $\xi$ and assume that $p : X \to Y$ is finite. Write $Y = \Spec(R)$ and $X = \Spec(A)$ with $p$ induced by a finite homomorphism $R \to A$ of Noetherian domains which induces an finite field extension $L/K$ of fraction fields. Now we have $f \in L$, $g = \text{Nm}(f) \in K$, and a prime $\mathfrak p \subset R$ with $\dim(R_{\mathfrak p}) = 1$. The coefficient of $[Z]$ in $\text{div}_Y(g)$ is $\text{ord}_{R_\mathfrak p}(g)$. The coefficient of $[Z]$ in $p_*\text{div}_X(f)$ is $$ \sum\nolimits_{\mathfrak q\text{ lying over }\mathfrak p} [\kappa(\mathfrak q) : \kappa(\mathfrak p)] \text{ord}_{A_{\mathfrak q}}(f) $$ The desired equality therefore follows from Algebra, Lemma \ref{algebra-lemma-finite-extension-dim-1}.
Since $X$ is integral, we see that $U$ is integral. Hence $Y$ is integral, and $(1, f)(U) \subset Y$ is an open dense subscheme. Also, note that the closed subscheme $Y \subset X \times_S \mathbf{P}^1_S$ does not depend on the choice of the open $U$, since after all it is the closure of the one point set $\{\eta'\} = \{(1, f)(\eta)\}$ where $\eta \in X$ is the generic point. Having said this let us prove the assertions of the lemma. \medskip\noindent For (1) note that $p$ is the composition of the closed immersion $Y \to X \times_S \mathbf{P}^1_S = \mathbf{P}^1_X$ with the proper morphism $\mathbf{P}^1_X \to X$. As a composition of proper morphisms is proper (Morphisms, Lemma \ref{morphisms-lemma-composition-proper}) we conclude. \medskip\noindent It is clear that $Y \cap U \times_S \mathbf{P}^1_S = (1, f)(U)$. Thus (2) follows. It also follows that $\dim_\delta(Y) = n$. \medskip\noindent Note that $q(\eta') = f(\eta)$ is not contained in $D_0$ or $D_\infty$ since $f \in R(X)^*$. Hence (3) by Divisors, Lemma \ref{divisors-lemma-pullback-effective-Cartier-defined}. We obtain $\dim_\delta(Y_0) = n - 1$ and $\dim_\delta(Y_\infty) = n - 1$ from Lemma \ref{lemma-divisor-delta-dimension}. \medskip\noindent Consider the effective Cartier divisor $Y_0$. At every point $\xi \in Y_0$ we have $f \in \mathcal{O}_{Y, \xi}$ and the local equation for $Y_0$ is given by $f$. In particular, if $\delta(\xi) = n - 1$ so $\xi$ is the generic point of a integral closed subscheme $Z$ of $\delta$-dimension $n - 1$, then we see that the coefficient of $[Z]$ in $\text{div}_Y(f)$ is $$ \text{ord}_Z(f) = \text{length}_{\mathcal{O}_{Y, \xi}} (\mathcal{O}_{Y, \xi}/f\mathcal{O}_{Y, \xi}) = \text{length}_{\mathcal{O}_{Y, \xi}} (\mathcal{O}_{Y_0, \xi}) $$ which is the coefficient of $[Z]$ in $[Y_0]_{n - 1}$. A similar argument using the rational function $1/f$ shows that $-[Y_\infty]$ agrees with the terms with negative coefficients in the expression for $\text{div}_Y(f)$. Hence (4) follows. \medskip\noindent Note that $D_0 \to S$ is an isomorphism. Hence we see that $X \times_S D_0 \to X$ is an isomorphism as well. Clearly we have $Y_0 = Y \cap X \times_S D_0$ (scheme theoretic intersection) inside $X \times_S \mathbf{P}^1_S$. Hence it is really the case that $Y_0 \to X$ is a closed immersion. It follows that $$ p_*\mathcal{O}_{Y_0} = \mathcal{O}_{Y'_0} $$ where $Y'_0 \subset X$ is the image of $Y_0 \to X$. By Lemma \ref{lemma-cycle-push-sheaf} we have $p_*[Y_0]_{n - 1} = [Y'_0]_{n - 1}$. The same is true for $D_\infty$ and $Y_\infty$. Hence (6) is a consequence of (5). Finally, (5) follows immediately from Lemma \ref{lemma-proper-pushforward-alteration}.
Consider the diagram $$ \xymatrix{ Y \ar[r]_p \ar[d]_q & X \ar[d]^c \\ \mathbf{P}^1_K \ar[r]^-{c'} & \Spec(K) } $$ that we constructed in Lemma \ref{lemma-rational-function} starting with $X$ and the rational function $f$ over $S = \Spec(K)$. We will use all the results of this lemma without further mention. We have to show that $c_*\text{div}_X(f) = c_*p_*\text{div}_Y(f) = 0$. This is the same as proving that $c'_*q_*\text{div}_Y(f) = 0$. If $q(Y)$ is a closed point of $\mathbf{P}^1_K$ then we see that $\text{div}_X(f) = 0$ and the lemma holds. Thus we may assume that $q$ is dominant. Suppose we can show that $q : Y \to \mathbf{P}^1_K$ is finite locally free of degree $d$ (see Morphisms, Definition \ref{morphisms-definition-finite-locally-free}). Since $\text{div}_Y(f) = [q^{-1}D_0]_0 - [q^{-1}D_\infty]_0$ we see (by definition of flat pullback) that $\text{div}_Y(f) = q^*([D_0]_0 - [D_\infty]_0)$. Then by Lemma \ref{lemma-finite-flat} we get $q_*\text{div}_Y(f) = d([D_0]_0 - [D_\infty]_0)$. Since clearly $c'_*[D_0]_0 = c'_*[D_\infty]_0$ we win. \medskip\noindent It remains to show that $q$ is finite locally free. (It will automatically have some given degree as $\mathbf{P}^1_K$ is connected.) Since $\dim(\mathbf{P}^1_K) = 1$ we see that $q$ is finite for example by Lemma \ref{lemma-finite-in-codimension-one}. All local rings of $\mathbf{P}^1_K$ at closed points are regular local rings of dimension $1$ (in other words discrete valuation rings), since they are localizations of $K[T]$ (see Algebra, Lemma \ref{algebra-lemma-dim-affine-space}). Hence for $y\in Y$ closed the local ring $\mathcal{O}_{Y, y}$ will be flat over $\mathcal{O}_{\mathbf{P}^1_K, q(y)}$ as soon as it is torsion free (More on Algebra, Lemma \ref{more-algebra-lemma-dedekind-torsion-free-flat}). This is obviously the case as $\mathcal{O}_{Y, y}$ is a domain and $q$ is dominant. Thus $q$ is flat. Hence $q$ is finite locally free by Morphisms, Lemma \ref{morphisms-lemma-finite-flat}.
Let $\{W_j\}_{j \in J}$ be a locally finite collection of integral closed subschemes of $U$ of $\delta$-dimension $k + 1$, and let $f_j \in R(W_j)^*$ be elements such that $(\alpha - \beta)|_U = \sum (i_j)_*\text{div}(f_j)$ as in the definition. Set $W_j' \subset X$ equal to the closure of $W_j$. Suppose that $V \subset X$ is a quasi-compact open. Then also $V \cap U$ is quasi-compact open in $U$ as $V$ is Noetherian. Hence the set $\{j \in J \mid W_j \cap V \not = \emptyset\} = \{j \in J \mid W'_j \cap V \not = \emptyset\}$ is finite since $\{W_j\}$ is locally finite. In other words we see that $\{W'_j\}$ is also locally finite. Since $R(W_j) = R(W'_j)$ we see that $$ \alpha - \beta - \sum (i'_j)_*\text{div}(f_j) $$ is a cycle supported on $Y$ and the lemma follows (see Lemma \ref{lemma-exact-sequence-open}).
By Lemma \ref{lemma-exact-sequence-closed} the arrow $\CH_k(X_1) \oplus \CH_k(X_2) \to \CH_k(X)$ is surjective. Suppose that $(\alpha_1, \alpha_2)$ maps to zero under this map. Write $\alpha_1 = \sum n_{1, i}[W_{1, i}]$ and $\alpha_2 = \sum n_{2, i}[W_{2, i}]$. Then we obtain a locally finite collection $\{W_j\}_{j \in J}$ of integral closed subschemes of $X$ of $\delta$-dimension $k + 1$ and $f_j \in R(W_j)^*$ such that $$ \sum n_{1, i}[W_{1, i}] + \sum n_{2, i}[W_{2, i}] = \sum (i_j)_*\text{div}(f_j) $$ as cycles on $X$ where $i_j : W_j \to X$ is the inclusion morphism. Choose a disjoint union decomposition $J = J_1 \amalg J_2$ such that $W_j \subset X_1$ if $j \in J_1$ and $W_j \subset X_2$ if $j \in J_2$. (This is possible because the $W_j$ are integral.) Then we can write the equation above as $$ \sum n_{1, i}[W_{1, i}] - \sum\nolimits_{j \in J_1} (i_j)_*\text{div}(f_j) = - \sum n_{2, i}[W_{2, i}] + \sum\nolimits_{j \in J_2} (i_j)_*\text{div}(f_j) $$ Hence this expression is a cycle (!) on $X_1 \cap X_2$. In other words the element $(\alpha_1, \alpha_2)$ is in the image of the first arrow and the proof is complete.
Let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k + r - 1$. We have to show that the coefficient $n$ of $[Z]$ in $f^*\text{div}(g)$ is equal to the coefficient $m$ of $[Z]$ in $\sum i_{j, *} \text{div}(g \circ f|_{X_j})$. Let $Z'$ be the closure of $f(Z)$ which is an integral closed subscheme of $Y$. By Lemma \ref{lemma-flat-inverse-image-dimension} we have $\dim_\delta(Z') \geq k - 1$. Thus either $Z' = Y$ or $Z'$ is a prime divisor on $Y$. If $Z' = Y$, then the coefficients $n$ and $m$ are both zero: this is clear for $n$ by definition of $f^*$ and follows for $m$ because $g \circ f|_{X_j}$ is a unit in any point of $X_j$ mapping to the generic point of $Y$. Hence we may assume that $Z' \subset Y$ is a prime divisor. \medskip\noindent We are going to translate the equality of $n$ and $m$ into algebra. Namely, let $\xi' \in Z'$ and $\xi \in Z$ be the generic points. Set $A = \mathcal{O}_{Y, \xi'}$ and $B = \mathcal{O}_{X, \xi}$. Note that $A$, $B$ are Noetherian, $A \to B$ is flat, local, $A$ is a domain, and $\mathfrak m_AB$ is an ideal of definition of the local ring $B$. The rational function $g$ is an element of the fraction field $Q(A)$ of $A$. By construction, the closed subschemes $X_j$ which meet $\xi$ correspond $1$-to-$1$ with minimal primes $$ \mathfrak q_1, \ldots, \mathfrak q_s \subset B $$ The integers $n_j$ are the corresponding lengths $$ n_i = \text{length}_{B_{\mathfrak q_i}}(B_{\mathfrak q_i}) $$ The rational functions $g \circ f|_{X_j}$ correspond to the image $g_i \in \kappa(\mathfrak q_i)^*$ of $g \in Q(A)$. Putting everything together we see that $$ n = \text{ord}_A(g) \text{length}_B(B/\mathfrak m_AB) $$ and that $$ m = \sum \text{ord}_{B/\mathfrak q_i}(g_i) \text{length}_{B_{\mathfrak q_i}}(B_{\mathfrak q_i}) $$ Writing $g = x/y$ for some nonzero $x, y \in A$ we see that it suffices to prove $$ \text{length}_A(A/(x)) \text{length}_B(B/\mathfrak m_AB) = \text{length}_B(B/xB) $$ (equality uses Algebra, Lemma \ref{algebra-lemma-pullback-module}) equals $$ \sum\nolimits_{i = 1, \ldots, s} \text{length}_{B/\mathfrak q_i}(B/(x, \mathfrak q_i)) \text{length}_{B_{\mathfrak q_i}}(B_{\mathfrak q_i}) $$ and similarly for $y$. As $A \to B$ is flat it follows that $x$ is a nonzerodivisor in $B$. Hence the desired equality follows from Lemma \ref{lemma-additivity-divisors-restricted}.
What do we have to show? Well, suppose we are given a collection $$ i_j : W_j \longrightarrow Y $$ of closed immersions, with each $W_j$ integral of $\delta$-dimension $k + 1$ and rational functions $g_j \in R(W_j)^*$. Moreover, assume that the collection $\{i_j(W_j)\}_{j \in J}$ is locally finite on $Y$. Then we have to show that $$ f^*(\sum i_{j, *}\text{div}(g_j)) = \sum f^*i_{j, *}\text{div}(g_j) $$ is rationally equivalent to zero on $X$. The sum on the right makes sense as $\{W_j\}$ is locally finite in $X$ by Lemma \ref{lemma-inverse-image-locally-finite}. \medskip\noindent Consider the fibre products $$ i'_j : W'_j = W_j \times_Y X \longrightarrow X. $$ and denote $f_j : W'_j \to W_j$ the first projection. By Lemma \ref{lemma-flat-pullback-proper-pushforward} we can write the sum above as $$ \sum i'_{j, *}(f_j^*\text{div}(g_j)) $$ By Lemma \ref{lemma-prepare-flat-pullback-rational-equivalence} we see that each $f_j^*\text{div}(g_j)$ is rationally equivalent to zero on $W'_j$. Hence each $i'_{j, *}(f_j^*\text{div}(g_j))$ is rationally equivalent to zero. Then the same is true for the displayed sum by the discussion in Remark \ref{remark-infinite-sums-rational-equivalences}.
What do we have to show? Well, suppose we are given a collection $$ i_j : W_j \longrightarrow X $$ of closed immersions, with each $W_j$ integral of $\delta$-dimension $k + 1$ and rational functions $f_j \in R(W_j)^*$. Moreover, assume that the collection $\{i_j(W_j)\}_{j \in J}$ is locally finite on $X$. Then we have to show that $$ p_*\left(\sum i_{j, *}\text{div}(f_j)\right) $$ is rationally equivalent to zero on $X$. \medskip\noindent Note that the sum is equal to $$ \sum p_*i_{j, *}\text{div}(f_j). $$ Let $W'_j \subset Y$ be the integral closed subscheme which is the image of $p \circ i_j$. The collection $\{W'_j\}$ is locally finite in $Y$ by Lemma \ref{lemma-quasi-compact-locally-finite}. Hence it suffices to show, for a given $j$, that either $p_*i_{j, *}\text{div}(f_j) = 0$ or that it is equal to $i'_{j, *}\text{div}(g_j)$ for some $g_j \in R(W'_j)^*$. \medskip\noindent The arguments above therefore reduce us to the case of a since integral closed subscheme $W \subset X$ of $\delta$-dimension $k + 1$. Let $f \in R(W)^*$. Let $W' = p(W)$ as above. We get a commutative diagram of morphisms $$ \xymatrix{ W \ar[r]_i \ar[d]_{p'} & X \ar[d]^p \\ W' \ar[r]^{i'} & Y } $$ Note that $p_*i_*\text{div}(f) = i'_*(p')_*\text{div}(f)$ by Lemma \ref{lemma-compose-pushforward}. As explained above we have to show that $(p')_*\text{div}(f)$ is the divisor of a rational function on $W'$ or zero. There are three cases to distinguish. \medskip\noindent The case $\dim_\delta(W') < k$. In this case automatically $(p')_*\text{div}(f) = 0$ and there is nothing to prove. \medskip\noindent The case $\dim_\delta(W') = k$. Let us show that $(p')_*\text{div}(f) = 0$ in this case. Let $\eta \in W'$ be the generic point. Note that $c : W_\eta \to \Spec(K)$ is a proper integral curve over $K = \kappa(\eta)$ whose function field $K(W_\eta)$ is identified with $R(W)$. Here is a diagram $$ \xymatrix{ W_\eta \ar[r] \ar[d]_c & W \ar[d]^{p'} \\ \Spec(K) \ar[r] & W' } $$ Let us denote $f_\eta \in K(W_\eta)^*$ the rational function corresponding to $f \in R(W)^*$. Moreover, the closed points $\xi$ of $W_\eta$ correspond $1 - 1$ to the closed integral subschemes $Z = Z_\xi \subset W$ of $\delta$-dimension $k$ with $p'(Z) = W'$. Note that the multiplicity of $Z_\xi$ in $\text{div}(f)$ is equal to $\text{ord}_{\mathcal{O}_{W_\eta, \xi}}(f_\eta)$ simply because the local rings $\mathcal{O}_{W_\eta, \xi}$ and $\mathcal{O}_{W, \xi}$ are identified (as subrings of their fraction fields). Hence we see that the multiplicity of $[W']$ in $(p')_*\text{div}(f)$ is equal to the multiplicity of $[\Spec(K)]$ in $c_*\text{div}(f_\eta)$. By Lemma \ref{lemma-curve-principal-divisor} this is zero. \medskip\noindent The case $\dim_\delta(W') = k + 1$. In this case Lemma \ref{lemma-proper-pushforward-alteration} applies, and we see that indeed $p'_*\text{div}(f) = \text{div}(g)$ for some $g \in R(W')^*$ as desired.
Part (1) follows from Divisors, Lemma \ref{divisors-lemma-pullback-effective-Cartier-defined} since the generic point of $W$ is not mapped into $D_0$ or $D_\infty$ under the projection $X \times_S \mathbf{P}^1_S \to \mathbf{P}^1_S$ by assumption. \medskip\noindent Since $X \times_S D_0 \to X$ is a closed immersion, we see that $W_0$ is isomorphic to a closed subscheme of $X$. Similarly for $W_\infty$. The morphism $p : W \to X$ is proper as a composition of the closed immersion $W \to X \times_S \mathbf{P}^1_S$ and the proper morphism $X \times_S \mathbf{P}^1_S \to X$. By Lemma \ref{lemma-rational-function} we have $[W_0]_k \sim_{rat} [W_\infty]_k$ as cycles on $W$. Hence part (2) follows from Lemma \ref{lemma-proper-pushforward-rational-equivalence} as clearly $p_*[W_0]_k = [W_0]_k$ and similarly for $W_\infty$. \medskip\noindent The only content of statement (3) is, given parts (1) and (2), that the collection $\{(W_i)_0, (W_i)_\infty\}$ is a locally finite collection of closed subschemes of $X$. This is clear. \medskip\noindent Suppose that $\alpha \sim_{rat} 0$. By definition this means there exist integral closed subschemes $V_i \subset X$ of $\delta$-dimension $k + 1$ and rational functions $f_i \in R(V_i)^*$ such that the family $\{V_i\}_{i \in I}$ is locally finite in $X$ and such that $\alpha = \sum (V_i \to X)_*\text{div}(f_i)$. Let $$ W_i \subset V_i \times_S \mathbf{P}^1_S \subset X \times_S \mathbf{P}^1_S $$ be the closure of the graph of the rational map $f_i$ as in Lemma \ref{lemma-rational-function}. Then we have that $(V_i \to X)_*\text{div}(f_i)$ is equal to $[(W_i)_0]_k - [(W_i)_\infty]_k$ by that same lemma. Hence the result is clear.
Let $\{W_i\}_{i \in I}$ be the collection of irreducible components of $Z$ which have $\delta$-dimension $k + 1$. Write $$ [Z]_{k + 1} = \sum n_i[W_i] $$ with $n_i > 0$ as per definition. Note that $\{W_i\}$ is a locally finite collection of closed subsets of $X \times_S \mathbf{P}^1_S$ by Divisors, Lemma \ref{divisors-lemma-components-locally-finite}. We claim that $$ [Z_0]_k = \sum n_i[(W_i)_0]_k $$ and similarly for $[Z_\infty]_k$. If we prove this then the lemma follows from Lemma \ref{lemma-rational-equivalence-family}. \medskip\noindent Let $Z' \subset X$ be an integral closed subscheme of $\delta$-dimension $k$. To prove the equality above it suffices to show that the coefficient $n$ of $[Z']$ in $[Z_0]_k$ is the same as the coefficient $m$ of $[Z']$ in $\sum n_i[(W_i)_0]_k$. Let $\xi' \in Z'$ be the generic point. Set $\xi = (\xi', 0) \in  X \times_S \mathbf{P}^1_S$. Consider the local ring $A = \mathcal{O}_{X \times_S \mathbf{P}^1_S, \xi}$. Let $I \subset A$ be the ideal cutting out $Z$, in other words so that $A/I = \mathcal{O}_{Z, \xi}$. Let $t \in A$ be the element cutting out $X \times_S D_0$ (i.e., the coordinate of $\mathbf{P}^1$ at zero pulled back). By our choice of $\xi' \in Z'$ we have $\delta(\xi) = k$ and hence $\dim(A/I) = 1$. Since $\xi$ is not an embedded point by assumption (3) we see that $A/I$ is Cohen-Macaulay. Since $\dim_\delta(Z_0) = k$ we see that $\dim(A/(t, I)) = 0$ which implies that $t$ is a nonzerodivisor on $A/I$. Finally, the irreducible closed subschemes $W_i$ passing through $\xi$ correspond to the minimal primes $I \subset \mathfrak q_i$ over $I$. The multiplicities $n_i$ correspond to the lengths $\text{length}_{A_{\mathfrak q_i}}(A/I)_{\mathfrak q_i}$. Hence we see that $$ n = \text{length}_A(A/(t, I)) $$ and $$ m = \sum \text{length}_A(A/(t, \mathfrak q_i)) \text{length}_{A_{\mathfrak q_i}}(A/I)_{\mathfrak q_i} $$ Thus the result follows from Lemma \ref{lemma-additivity-divisors-restricted}.
Let $\{W_i\}_{i \in I}$ be the collection of irreducible components of $\text{Supp}(\mathcal{F})$ which have $\delta$-dimension $k + 1$. Write $$ [\mathcal{F}]_{k + 1} = \sum n_i[W_i] $$ with $n_i > 0$ as per definition. Note that $\{W_i\}$ is a locally finite collection of closed subsets of $X \times_S \mathbf{P}^1_S$ by Lemma \ref{lemma-length-finite}. We claim that $$ [\mathcal{F}_0]_k = \sum n_i[(W_i)_0]_k $$ and similarly for $[\mathcal{F}_\infty]_k$. If we prove this then the lemma follows from Lemma \ref{lemma-rational-equivalence-family}. \medskip\noindent Let $Z' \subset X$ be an integral closed subscheme of $\delta$-dimension $k$. To prove the equality above it suffices to show that the coefficient $n$ of $[Z']$ in $[\mathcal{F}_0]_k$ is the same as the coefficient $m$ of $[Z']$ in $\sum n_i[(W_i)_0]_k$. Let $\xi' \in Z'$ be the generic point. Set $\xi = (\xi', 0) \in  X \times_S \mathbf{P}^1_S$. Consider the local ring $A = \mathcal{O}_{X \times_S \mathbf{P}^1_S, \xi}$. Let $M = \mathcal{F}_\xi$ as an $A$-module. Let $t \in A$ be the element cutting out $X \times_S D_0$ (i.e., the coordinate of $\mathbf{P}^1$ at zero pulled back). By our choice of $\xi' \in Z'$ we have $\delta(\xi) = k$ and hence $\dim(\text{Supp}(M)) = 1$. Since $\xi$ is not an associated point of $\mathcal{F}$ by assumption (3) we see that $M$ is a Cohen-Macaulay module. Since $\dim_\delta(\text{Supp}(\mathcal{F}_0)) = k$ we see that $\dim(\text{Supp}(M/tM)) = 0$ which implies that $t$ is a nonzerodivisor on $M$. Finally, the irreducible closed subschemes $W_i$ passing through $\xi$ correspond to the minimal primes $\mathfrak q_i$ of $\text{Ass}(M)$. The multiplicities $n_i$ correspond to the lengths $\text{length}_{A_{\mathfrak q_i}}M_{\mathfrak q_i}$. Hence we see that $$ n = \text{length}_A(M/tM) $$ and $$ m = \sum \text{length}_A(A/(t, \mathfrak q_i)A) \text{length}_{A_{\mathfrak q_i}}M_{\mathfrak q_i} $$ Thus the result follows from Lemma \ref{lemma-additivity-divisors-restricted}.
A finite morphism is affine, hence pushforward of quasi-coherent modules along $\pi$ is an exact functor by Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-vanishing}. A finite morphism is proper, hence $\pi_*$ sends coherent sheaves to coherent sheaves, see Cohomology of Schemes, Proposition \ref{coherent-proposition-proper-pushforward-coherent}. The statement on dimensions of supports is clear. Commutativity on the right follows immediately from Lemma \ref{lemma-cycle-push-sheaf}. Since the horizontal arrows are bijections, we find that we have commutativity on the left as well.
We have to show that an element $\alpha$ of $Z_k(X)$ which is rationally equivalent to zero, is mapped to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. Write $\alpha = \sum (i_j)_*\text{div}(f_j)$ as in Definition \ref{definition-rational-equivalence}. Observe that $$ \pi = \coprod i_j : W = \coprod W_j \longrightarrow X $$ is a finite morphism as each $i_j : W_j \to X$ is a closed immersion and the family of $W_j$ is locally finite in $X$. Hence we may use Lemma \ref{lemma-finite-cycles-k-group} to reduce to the case of $W$. Since $W$ is a disjoint union of intregral scheme, we reduce to the case discussed in the next paragraph. \medskip\noindent Assume $X$ is integral of $\delta$-dimension $k + 1$. Let $f$ be a nonzero rational function on $X$. Let $\alpha = \text{div}(f)$. We have to show that $\alpha$ is mapped to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. Let $\mathcal{I} \subset \mathcal{O}_X$ be the ideal of denominators of $f$, see Divisors, Definition \ref{divisors-definition-regular-meromorphic-ideal-denominators}. Then we have short exact sequences $$ 0 \to \mathcal{I} \to \mathcal{O}_X \to \mathcal{O}_X/\mathcal{I} \to 0 $$ and $$ 0 \to \mathcal{I} \xrightarrow{f} \mathcal{O}_X \to \mathcal{O}_X/f\mathcal{I} \to 0 $$ See Divisors, Lemma \ref{divisors-lemma-regular-meromorphic-ideal-denominators}. We claim that $$ [\mathcal{O}_X/\mathcal{I}]_k - [\mathcal{O}_X/f\mathcal{I}]_k = \text{div}(f) $$ The claim implies the element $\alpha = \text{div}(f)$ is represented by $[\mathcal{O}_X/\mathcal{I}] - [\mathcal{O}_X/f\mathcal{I}]$ in $K_0(\textit{Coh}_{\leq k}(X)/\textit{Coh}_{\leq k - 1}(X))$. Then the short exact sequences show that this element maps to zero in $K_0(\textit{Coh}_{\leq k + 1}(X)/\textit{Coh}_{\leq k - 1}(X))$. \medskip\noindent To prove the claim, let $Z \subset X$ be an integral closed subscheme of $\delta$-dimension $k$ and let $\xi \in Z$ be its generic point. Then $I = \mathcal{I}_\xi \subset A = \mathcal{O}_{X, \xi}$ is an ideal such that $fI \subset A$. Now the coefficient of $[Z]$ in $\text{div}(f)$ is $\text{ord}_A(f)$. (Of course as usual we identify the function field of $X$ with the fraction field of $A$.) On the other hand, the coefficient of $[Z]$ in $[\mathcal{O}_X/\mathcal{I}] - [\mathcal{O}_X/f\mathcal{I}]$ is $$ \text{length}_A(A/I) - \text{length}_A(A/fI) $$ Using the distance fuction of Algebra, Definition \ref{algebra-definition-distance} we can rewrite this as $$ d(A, I) - d(A, fI) = d(I, fI) = \text{ord}_A(f) $$ The equalities hold by Algebra, Lemmas \ref{algebra-lemma-properties-distance-function} and \ref{algebra-lemma-order-vanishing-determinant}. (Using these lemmas isn't necessary, but convenient.)