## 42.20 Rational equivalence and push and pull

In this section we show that flat pullback and proper pushforward commute with rational equivalence.

Lemma 42.20.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be schemes locally of finite type over $S$. Assume $Y$ integral with $\dim _\delta (Y) = k$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Then for $g \in R(Y)^*$ we have

\[ f^*\text{div}_ Y(g) = \sum n_ j i_{j, *}\text{div}_{X_ j}(g \circ f|_{X_ j}) \]

as $(k + r - 1)$-cycles on $X$ where the sum is over the irreducible components $X_ j$ of $X$ and $n_ j$ is the multiplicity of $X_ j$ in $X$.

**Proof.**
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 42.13.1 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.

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 10.52.13) 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 42.3.2.
$\square$

Lemma 42.20.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be schemes locally of finite type over $S$. Let $f : X \to Y$ be a flat morphism of relative dimension $r$. Let $\alpha \sim _{rat} \beta $ be rationally equivalent $k$-cycles on $Y$. Then $f^*\alpha \sim _{rat} f^*\beta $ as $(k + r)$-cycles on $X$.

**Proof.**
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 42.13.2.

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 42.15.1 we can write the sum above as

\[ \sum i'_{j, *}(f_ j^*\text{div}(g_ j)) \]

By Lemma 42.20.1 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 42.19.6.
$\square$

Lemma 42.20.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$, $Y$ be schemes locally of finite type over $S$. Let $p : X \to Y$ be a proper morphism. Suppose $\alpha , \beta \in Z_ k(X)$ are rationally equivalent. Then $p_*\alpha $ is rationally equivalent to $p_*\beta $.

**Proof.**
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 $X$.

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 42.11.2. 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 $g_ j \in R(W'_ j)^*$.

The arguments above therefore reduce us to the case of a single 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 42.12.2. 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.

The case $\dim _\delta (W') < k$. In this case automatically $(p')_*\text{div}(f) = 0$ and there is nothing to prove.

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 $c : W_\eta \to \mathop{\mathrm{Spec}}(K)$ is a proper integral curve over $K = \kappa (\eta )$ whose function field $K(W_\eta )$ is identified with $R(W)$. Here is a diagram

\[ \xymatrix{ W_\eta \ar[r] \ar[d]_ c & W \ar[d]^{p'} \\ \mathop{\mathrm{Spec}}(K) \ar[r] & W' } \]

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 $[\mathop{\mathrm{Spec}}(K)]$ in $c_*\text{div}(f_\eta )$. By Lemma 42.18.3 this is zero.

The case $\dim _\delta (W') = k + 1$. In this case Lemma 42.18.1 applies, and we see that indeed $p'_*\text{div}(f) = \text{div}(g)$ for some $g \in R(W')^*$ as desired.
$\square$

## Comments (2)

Comment #7774 by Danny on

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