The Stacks project

42.29 Gysin homomorphisms and rational equivalence

In this section we use the key formula to show the Gysin homomorphism factor through rational equivalence. We also prove an important commutativity property.

Lemma 42.29.1. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $X$ be integral and $n = \dim _\delta (X)$. Let $i : D \to X$ be an effective Cartier divisor. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Then $i^*\text{div}_\mathcal {N}(t) = c_1(\mathcal{N}) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$.

Proof. Write $\text{div}_\mathcal {N}(t) = \sum \text{ord}_{Z_ i, \mathcal{N}}(t)[Z_ i]$ for some integral closed subschemes $Z_ i \subset X$ of $\delta $-dimension $n - 1$. We may assume that the family $\{ Z_ i\} $ is locally finite, that $t \in \Gamma (U, \mathcal{N}|_ U)$ is a generator where $U = X \setminus \bigcup Z_ i$, and that every irreducible component of $D$ is one of the $Z_ i$, see Divisors, Lemmas 31.26.1, 31.26.4, and 31.27.2.

Set $\mathcal{L} = \mathcal{O}_ X(D)$. Denote $s \in \Gamma (X, \mathcal{O}_ X(D)) = \Gamma (X, \mathcal{L})$ the canonical section. We will apply the discussion of Section 42.26 to our current situation. For each $i$ let $\xi _ i \in Z_ i$ be its generic point. Let $B_ i = \mathcal{O}_{X, \xi _ i}$. For each $i$ we pick generators $s_ i \in \mathcal{L}_{\xi _ i}$ and $t_ i \in \mathcal{N}_{\xi _ i}$ over $B_ i$ but we insist that we pick $s_ i = s$ if $Z_ i \not\subset D$. Write $s = f_ i s_ i$ and $t = g_ i t_ i$ with $f_ i, g_ i \in B_ i$. Then $\text{ord}_{Z_ i, \mathcal{N}}(t) = \text{ord}_{B_ i}(g_ i)$. On the other hand, we have $f_ i \in B_ i$ and

\[ [D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i)[Z_ i] \]

because of our choices of $s_ i$. We claim that

\[ i^*\text{div}_\mathcal {N}(t) = \sum \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \]

as cycles. More precisely, the right hand side is a cycle representing the left hand side. Namely, this is clear by our formula for $\text{div}_\mathcal {N}(t)$ and the fact that $\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) = [Z(s_ i|_{Z_ i})]_{n - 2} = [Z_ i \cap D]_{n - 2}$ when $Z_ i \not\subset D$ because in that case $s_ i|_{Z_ i} = s|_{Z_ i}$ is a regular section, see Lemma 42.23.2. Similarly,

\[ c_1(\mathcal{N}) \cap [D]_{n - 1} = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) \]

The key formula (Lemma 42.26.1) gives the equality

\[ \sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i)) \]

of cycles. If $Z_ i \not\subset D$, then $f_ i = 1$ and hence $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, g_ i)) = 0$. Thus we get a rational equivalence between our specific cycles representing $i^*\text{div}_\mathcal {N}(t)$ and $c_1(\mathcal{N}) \cap [D]_{n - 1}$ on $D$. This finishes the proof. $\square$

Lemma 42.29.2. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ be as in Definition 42.28.1. The Gysin homomorphism factors through rational equivalence to give a map $i^* : \mathop{\mathrm{CH}}\nolimits _{k + 1}(X) \to \mathop{\mathrm{CH}}\nolimits _ k(D)$.

Proof. Let $\alpha \in Z_{k + 1}(X)$ and assume that $\alpha \sim _{rat} 0$. This means there exists a locally finite collection of integral closed subschemes $W_ j \subset X$ of $\delta $-dimension $k + 2$ and $f_ j \in R(W_ j)^*$ such that $\alpha = \sum i_{j, *}\text{div}_{W_ j}(f_ j)$. Set $X' = \coprod W_ i$ and consider the diagram

\[ \xymatrix{ D' \ar[d]_ q \ar[r]_{i'} & X' \ar[d]^ p \\ D \ar[r]^ i & X } \]

of Remark 42.28.7. Since $X' \to X$ is proper we see that $i^*p_* = q_*(i')^*$ by Lemma 42.28.8. As we know that $q_*$ factors through rational equivalence (Lemma 42.20.3), it suffices to prove the result for $\alpha ' = \sum \text{div}_{W_ j}(f_ j)$ on $X'$. Clearly this reduces us to the case where $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$.

Assume $X$ is integral and $\alpha = \text{div}(f)$ for some $f \in R(X)^*$. If $X = D$, then we see that $i^*\alpha $ is equal to $c_1(\mathcal{L}) \cap \alpha $. This is rationally equivalent to zero by Lemma 42.27.2. If $D \not= X$, then we see that $i^*\text{div}_ X(f)$ is equal to $c_1(\mathcal{O}_ D) \cap [D]_{n - 1}$ in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(D)$ by Lemma 42.29.1. Of course capping with $c_1(\mathcal{O}_ D)$ is the zero map (Lemma 42.24.2). $\square$

Lemma 42.29.3. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ be as in Definition 42.28.1. Then $i^*i_* : \mathop{\mathrm{CH}}\nolimits _ k(X) \to \mathop{\mathrm{CH}}\nolimits _{k - 1}(X)$ sends $\alpha $ to $c_1(\mathcal{L}|_ D) \cap \alpha $.

Proof. This is immediate from the definition of $i_*$ on cycles and the definition of $i^*$ given in Definition 42.28.1. $\square$

Lemma 42.29.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ be a triple as in Definition 42.28.1. Let $\mathcal{N}$ be an invertible $\mathcal{O}_ X$-module. Then $i^*(c_1(\mathcal{N}) \cap \alpha ) = c_1(i^*\mathcal{N}) \cap i^*\alpha $ in $\mathop{\mathrm{CH}}\nolimits _{k - 2}(D)$ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(Z)$.

Lemma 42.29.5. Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Let $(\mathcal{L}, s, i : D \to X)$ and $(\mathcal{L}', s', i' : D' \to X)$ be two triples as in Definition 42.28.1. Then the diagram

\[ \xymatrix{ \mathop{\mathrm{CH}}\nolimits _ k(X) \ar[r]_{i^*} \ar[d]_{(i')^*} & \mathop{\mathrm{CH}}\nolimits _{k - 1}(D) \ar[d]^{j^*} \\ \mathop{\mathrm{CH}}\nolimits _{k - 1}(D') \ar[r]^{(j')^*} & \mathop{\mathrm{CH}}\nolimits _{k - 2}(D \cap D') } \]

commutes where each of the maps is a gysin map.

Proof. Denote $j : D \cap D' \to D$ and $j' : D \cap D' \to D'$ the closed immersions corresponding to $(\mathcal{L}|_{D'}, s|_{D'}$ and $(\mathcal{L}'_ D, s|_ D)$. We have to show that $(j')^*i^*\alpha = j^* (i')^*\alpha $ for all $\alpha \in \mathop{\mathrm{CH}}\nolimits _ k(X)$. Let $W \subset X$ be an integral closed subscheme of dimension $k$. Let us prove the equality in case $\alpha = [W]$. We will deduce it from the key formula.

We let $\sigma $ be a nonzero meromorphic section of $\mathcal{L}|_ W$ which we require to be equal to $s|_ W$ if $W \not\subset D$. We let $\sigma '$ be a nonzero meromorphic section of $\mathcal{L}'|_ W$ which we require to be equal to $s'|_ W$ if $W \not\subset D'$. Write

\[ \text{div}_{\mathcal{L}|_ W}(\sigma ) = \sum \text{ord}_{Z_ i, \mathcal{L}|_ W}(\sigma )[Z_ i] = \sum n_ i[Z_ i] \]

and similarly

\[ \text{div}_{\mathcal{L}'|_ W}(\sigma ') = \sum \text{ord}_{Z_ i, \mathcal{L}'|_ W}(\sigma ')[Z_ i] = \sum n'_ i[Z_ i] \]

as in the discussion in Section 42.26. Then we see that $Z_ i \subset D$ if $n_ i \not= 0$ and $Z'_ i \subset D'$ if $n'_ i \not= 0$. For each $i$, let $\xi _ i \in Z_ i$ be the generic point. As in Section 42.26 we choose for each $i$ an element $\sigma _ i \in \mathcal{L}_{\xi _ i}$, resp. $\sigma '_ i \in \mathcal{L}'_{\xi _ i}$ which generates over $B_ i = \mathcal{O}_{W, \xi _ i}$ and which is equal to the image of $s$, resp. $s'$ if $Z_ i \not\subset D$, resp. $Z_ i \not\subset D'$. Write $\sigma = f_ i \sigma _ i$ and $\sigma ' = f'_ i\sigma '_ i$ so that $n_ i = \text{ord}_{B_ i}(f_ i)$ and $n'_ i = \text{ord}_{B_ i}(f'_ i)$. From our definitions it follows that

\[ (j')^*i^*[W] = \sum \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i}) \]

as cycles and

\[ j^*(i')^*[W] = \sum \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i}) \]

The key formula (Lemma 42.26.1) now gives the equality

\[ \sum \left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{L}'|_{Z_ i}}(\sigma '_ i|_{Z_ i}) - \text{ord}_{B_ i}(f'_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(\sigma _ i|_{Z_ i}) \right) = \sum \text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i)) \]

of cycles. Note that $\text{div}_{Z_ i}(\partial _{B_ i}(f_ i, f'_ i)) = 0$ if $Z_ i \not\subset D \cap D'$ because in this case either $f_ i = 1$ or $f'_ i = 1$. Thus we get a rational equivalence between our specific cycles representing $(j')^*i^*[W]$ and $j^*(i')^*[W]$ on $D \cap D' \cap W$. By Remark 42.19.5 the result follows for general $\alpha $. $\square$


Comments (2)

Comment #2983 by Xia on

In line 7 of the proof of the first lemma, the latter L should be N

Comment #3107 by on

THanks, fixed here. Please leave comments on lemmas on the page of the lemma.


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