The Stacks project

Proof. The projection $X' \to X$ is flat as a base change of the flat morphism $S' \to S$ (Morphisms, Lemma 29.25.8). Hence every generic point $x'$ of an irreducible component of $X'$ maps to the generic point $x \in X$ (because generalizations lift along $X' \to X$ by Morphisms, Lemma 29.25.9). Let $s \in S$ be the image of $x$. Recall that the scheme $S'_ s = S' \times _ S s$ has the same underlying topological space as $g^{-1}(\{ s\} )$ (Schemes, Lemma 26.18.5). We may view $x'$ as a point of the scheme $S'_ s \times _ s x$ which comes equipped with a monomorphism $S'_ s \times _ s x \to S' \times _ S X$. Of course, $x'$ is a generic point of an irreducible component of $S'_ s \times _ s x$ as well. Using the flatness of $\mathop{\mathrm{Spec}}(\kappa (x)) \to \mathop{\mathrm{Spec}}(\kappa (s)) = s$ and arguing as above, we see that $x'$ maps to a generic point $s'$ of an irreducible component of $g^{-1}(\{ s\} )$. Hence $\delta '(s') = \delta (s) + c$ by assumption. We have $\dim _ x(X_ s) = \dim _{x'}(X_{s'})$ by Morphisms, Lemma 29.28.3. Since $x$ is a generic point of an irreducible component $X_ s$ (this is an irreducible scheme but we don't need this) and $x'$ is a generic point of an irreducible component of $X'_{s'}$ we conclude that $\text{trdeg}_{\kappa (s)}(\kappa (x)) = \text{trdeg}_{\kappa (s')}(\kappa (x'))$ by Morphisms, Lemma 29.28.1. Then

This proves what we want by Definition 42.7.6. $\square$

Proof. The proof is exactly the same is the proof of Lemma 42.14.4 and we suggest the reader skip it.

The statements on dimensions follow from Lemma 42.67.2. Part (1) follows from part (2) by Lemma 42.10.3 and the fact that the base change of the coherent module $\mathcal{O}_ Z$ is $\mathcal{O}_{Z'}$.

Proof of (2). As $X$, $X'$ are locally Noetherian we may apply Cohomology of Schemes, Lemma 30.9.1 to see that $\mathcal{F}$ is of finite type, hence $\mathcal{F}'$ is of finite type (Modules, Lemma 17.9.2), hence $\mathcal{F}'$ is coherent (Cohomology of Schemes, Lemma 30.9.1 again). Thus the lemma makes sense. Let $W \subset X$ be an integral closed subscheme of $\delta $-dimension $k$, and let $W' \subset X'$ be an integral closed subscheme of $\delta '$-dimension $k + c$ mapping into $W$ under $X' \to X$. We have to show that the coefficient $n$ of $[W']$ in $g^*[\mathcal{F}]_ k$ agrees with the coefficient $m$ of $[W']$ in $[\mathcal{F}']_{k + c}$. Let $\xi \in W$ and $\xi ' \in W'$ be the generic points. Let $A = \mathcal{O}_{X, \xi }$, $B = \mathcal{O}_{X', \xi '}$ and set $M = \mathcal{F}_\xi $ as an $A$-module. (Note that $M$ has finite length by our dimension assumptions, but we actually do not need to verify this. See Lemma 42.10.1.) We have $\mathcal{F}'_{\xi '} = B \otimes _ A M$. Thus we see that

Thus the equality follows from Algebra, Lemma 10.52.13. $\square$

Proof. Suppose that $\alpha \in Z_ k(X)$ is a $k$-cycle which is rationally equivalent to zero. By Lemma 42.21.1 there exists a locally finite family of integral closed subschemes $W_ i \subset X \times \mathbf{P}^1$ of $\delta $-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty $ of $X \times \mathbf{P}^1$ such that $\alpha = \sum ([(W_ i)_0]_ k - [(W_ i)_\infty ]_ k)$. Thus it suffices to prove for $W \subset X \times \mathbf{P}^1$ integral closed of $\delta $-dimension $k$ not contained in the divisors $(X \times \mathbf{P}^1)_0$ or $(X \times \mathbf{P}^1)_\infty $ of $X \times \mathbf{P}^1$ we have

1. the base change $W' \subset X' \times \mathbf{P}^1$ satisfies the assumptions of Lemma 42.21.2 with $k$ replaced by $k + c$, and

2. $g^*[W_0]_ k = [(W')_0]_{k + c}$ and $g^*[W_\infty ]_ k = [(W')_\infty ]_{k + c}$.

Part (2) follows immediately from Lemma 42.67.3 and the fact that $(W')_0$ is the base change of $W_0$ (by associativity of fibre products). For part (1), first the statement on dimensions follows from Lemma 42.67.2. Then let $w' \in (W')_0$ with image $w \in W_0$ and $z \in \mathbf{P}^1_ S$. Denote $t \in \mathcal{O}_{\mathbf{P}^1_ S, z}$ the usual equation for $0 : S \to \mathbf{P}^1_ S$. Since $\mathcal{O}_{W, w} \to \mathcal{O}_{W', w'}$ is flat and since $t$ is a nonzerodivisor on $\mathcal{O}_{W, w}$ (as $W$ is integral and $W \not= W_0$) we see that also $t$ is a nonzerodivisor in $\mathcal{O}_{W', w'}$. Hence $W'$ has no associated points lying on $W'_0$. $\square$

Proof. It suffices to show the first diagram commutes. To see this, let $Z \subset X$ be an integral closed subscheme of $\delta $-dimension $k$ and denote $Z' \subset X'$ its base change. By construction we have $g^*[Z] = [Z']_{k + c}$. By Lemma 42.14.4 we have $(f')^*g^*[Z] = [Z' \times _{X'} Y']_{k + c + r}$. Conversely, we have $f^*[Z] = [Z \times _ X Y]_{k + r}$ by Definition 42.14.1. By Lemma 42.67.3 we have $g^*f^*[Z] = [(Z \times _ X Y)']_{k + r + c}$. Since $(Z \times _ X Y)' = Z' \times _{X'} Y'$ by associativity of fibre product we conclude. $\square$

Proof. It suffices to show the first diagram commutes. To see this, let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $k$ and denote $Z' \subset X'$ its base change. By construction we have $g^*[Z] = [Z']_{k + c}$. By Lemma 42.12.4 we have $(f')_*g^*[Z] = [f'_*\mathcal{O}_{Z'}]_{k + c}$. By the same lemma we have $f_*[Z] = [f_*\mathcal{O}_ Z]_ k$. By Lemma 42.67.3 we have $g^*f_*[Z] = [(X' \to X)^*f_*\mathcal{O}_ Z]_{k + r}$. Thus it suffices to show that

as coherent modules on $X'$. As $X' \to X$ is flat and as $\mathcal{O}_{Z'} = (Y' \to Y)^*\mathcal{O}_ Z$, this follows from flat base change, see Cohomology of Schemes, Lemma 30.5.2. $\square$

Proof. Let $p : L \to X$ be the line bundle associated to $\mathcal{L}$ with zero section $o : X \to L$. For $\alpha \in CH_ k(X)$ we know that $\beta = c_1(\mathcal{L}) \cap \alpha $ is the unique element of $\mathop{\mathrm{CH}}\nolimits _{k - 1}(X)$ such that $o_*\alpha = - p^*\beta $, see Lemmas 42.32.2 and 42.32.4. The same characterization holds after pullback. Hence the lemma follows from Lemmas 42.67.5 and 42.67.6. $\square$

Proof. Set $P = \mathbf{P}(\mathcal{E})$. The base change $P'$ of $P$ is equal to $\mathbf{P}(\mathcal{E}')$. Since we already know that flat pullback and cupping with $c_1$ of an invertible module commute with base change (Lemmas 42.67.5 and 42.67.7) the lemma follows from the characterization of capping with $c_ i(\mathcal{E})$ given in Lemma 42.38.2. $\square$

Proof. Let $s \in S$ and let $s'' \in S''$ be a generic point of an irreducible component of $(g \circ g')^{-1}(\{ s\} )$. Set $s' = g'(s'')$. Clearly, $s''$ is a generic point of an irreducible component of $(g')^{-1}(\{ s'\} )$. Moreover, since $g'$ is flat and hence generalizations lift along $g'$ (Morphisms, Lemma 29.25.8) we see that also $s'$ is a generic point of an irreducible component of $g^{-1}(\{ s\} )$. Thus by assumption $\delta '(s') = \delta (s) + c$ and $\delta ''(s'') = \delta '(s') + c'$. We conclude $\delta ''(s'') = \delta (s) + c + c'$ and the first part of the statement is true.

For the second part, let $Z \subset X$ be an integral closed subscheme of $\delta $-dimension $k$. Denote $Z' \subset X'$ and $Z'' \subset X''$ the base changes. By definition we have $g^*[Z] = [Z']_{k + c}$. By Lemma 42.67.3 we have $(g')^*[Z']_{k + c} = [Z'']_{k + c + c'}$. This proves the final statement. $\square$

Proof. By the result of Lemma 42.67.9 we obtain a system of cycle groups $Z_ k(X_ i)$ and a system of chow groups $\mathop{\mathrm{CH}}\nolimits _ k(X_ i)$ as well as maps $\mathop{\mathrm{colim}}\nolimits Z_ k(X_ i) \to Z_ k(X')$ and $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{CH}}\nolimits _ i(X_ i) \to \mathop{\mathrm{CH}}\nolimits _ k(X')$. We may replace $S$ by a quasi-compact open through which $X \to S$ factors, hence we may and do assume all the schemes occuring in this proof are Noetherian (and hence quasi-compact and quasi-separated).

Let us show that the map $\mathop{\mathrm{colim}}\nolimits Z_ k(X_ i) \to Z_ k(X')$ is surjective. Namely, let $Z' \subset X'$ be an integral closed subscheme of $\delta '$-dimension $k$. By Limits, Lemma 32.10.1 we can find an $i$ and a morphism $Z_ i \to X_ i$ of finite presentation whose base change is $Z'$. Afer increasing $i$ we may assume $Z_ i$ is a closed subscheme of $X_ i$, see Limits, Lemma 32.8.5. Then $Z' \to X_ i$ factors through $Z_ i$ and we may replace $Z_ i$ by the scheme theoretic image of $Z' \to X_ i$. In this way we see that we may assume $Z_ i$ is an integral closed subscheme of $X_ i$. By Lemma 42.67.2 we conclude that $\dim _{\delta _ i}(Z_ i) = \dim _{\delta '}(Z') = k$. Thus $Z_ k(X_ i) \to Z_ k(X')$ maps $[Z_ i]$ to $[Z']$ and we conclude surjectivity holds.

Let us show that the map $\mathop{\mathrm{colim}}\nolimits Z_ k(X_ i) \to Z_ k(X')$ is injective. Let $\alpha _ i = \sum n_ j[Z_ j] \in Z_ k(X_ i)$ be a cycle whose image in $Z_ k(X')$ is zero. We may and do assume $Z_ j \not= Z_{j'}$ if $j \not= j'$ and $n_ j \not= 0$ for all $j$. Denote $Z'_ j \subset X'$ the base change of $Z_ j$. By Lemma 42.67.2 each irreducible component of $Z'_ j$ has $\delta '$-dimension $k$. Moreover, as $Z_ j$ is irreducible and $Z'_ j \to Z_ j$ is flat (as the base change of $S' \to S$) we see that $Z'_ j \to Z_ j$ is dominant. Hence if $Z'_ j$ is nonempty, then some irreducible component, say $Z'$, of $Z'_ j$ dominates $Z_ j$. It follows that $Z'$ cannot be an irreducible component of $Z'_{j'}$ for $j' \not= j$. Hence if $Z'_ j$ is nonempty, then we see that $(S' \to S_ i)^*\alpha _ i = \sum [Z'_ j]_ r$ is nonzero (as the coefficient of $Z'$ would be nonzero). Thus we see that $Z'_ j = \emptyset $ for all $j$. However, this means that the base change of $Z_ j$ by some transition map $S_{i'} \to S_ i$ is empty by Limits, Lemma 32.4.3. Thus $\alpha _ i$ dies in the colimit as desired.

The surjectivity of $\mathop{\mathrm{colim}}\nolimits Z_ k(X_ i) \to Z_ k(X')$ implies that $\mathop{\mathrm{colim}}\nolimits \mathop{\mathrm{CH}}\nolimits _ k(X_ i) \to \mathop{\mathrm{CH}}\nolimits _ k(X')$ is surjective. To finish the proof we show that this map is injective. Let $\alpha _ i \in \mathop{\mathrm{CH}}\nolimits _ k(X_ i)$ be a cycle whose image $\alpha ' \in \mathop{\mathrm{CH}}\nolimits _ k(X')$ is zero. Then there exist integral closed subschemes $W_ l' \subset X'$, $l = 1, \ldots , r$ of $\delta "$-dimension $k + 1$ and nonzero rational functions $f'_ l$ on $W'_ l$ such that $\alpha ' = \sum _{l = 1, \ldots , r} \text{div}_{W'_ l}(f'_ l)$. Arguing as above we can find an $i$ and integral closed subschemes $W_{i, l} \subset X_ i$ of $\delta _ i$-dimension $k + 1$ whose base change is $W'_ l$. After increasin $i$ we may assume we have rational functions $f_{i, l}$ on $W_{i, l}$. Namely, we may think of $f'_ l$ as a section of the structure sheaf over a nonempty open $U'_ l \subset W'_ l$, we can descend these opens by Limits, Lemma 32.4.11 and after increasing $i$ we may descend $f'_ l$ by Limits, Lemma 32.4.7. We claim that

after possibly increasing $i$.

To prove the claim, let $Z'_{l, j} \subset W'_ l$ be a finite collection of integral closed subschemes of $\delta '$-dimension $k$ such that $f'_ l$ is an invertible regular function outside $\bigcup _ j Y'_{l, j}$. After increasing $i$ (by the arguments above) we may assume there exist integral closed subschemes $Z_{i, l, j} \subset W_ i$ of $\delta _ i$-dimension $k$ such that $f_{i, l}$ is an invertible regular function outside $\bigcup _ j Z_{i, l, j}$. Then we may write

and

To prove the claim it suffices to show that $n_{l, i} = n_{i, l, j}$. Namely, this will imply that $\beta _ i = \alpha _ i - \sum \nolimits _{l = 1, \ldots , r} \text{div}_{W_{i, l}}(f_{i, l})$ is a cycle on $X_ i$ whose pullback to $X'$ is zero as a cycle! It follows that $\beta _ i$ pulls back to zero as a cycle on $X_{i'}$ for some $i' \geq i$ by an easy argument we omit.

To prove the equality $n_{l, i} = n_{i, l, j}$ we choose a generic point $\xi ' \in Z'_{l, j}$ and we denote $\xi \in Z_{i, l, j}$ the image which is a generic point also. Then the local ring map

is flat as $W'_ l \to W_{i, l}$ is the base change of the flat morphism $S' \to S_ i$. We also have $\mathfrak m_\xi \mathcal{O}_{W'_ l, \xi '} = \mathfrak m_{\xi '}$ because $Z_{i, l, j}$ pulls back to $Z'_{l, j}$! Thus the equality of

follows from Algebra, Lemma 10.52.13 and the construction of $\text{ord}$ in Algebra, Section 10.121. $\square$