The Stacks project

Proof. For flat pullback use Lemma 62.4.4. Restriction is a special case of flat pullback. To see it holds for base change use that base change is transitive. For proper pushforward use Lemma 62.4.5. $\square$

Proof. If $\alpha $ is a relative $r$-cycle, then each base change $g_ i^*\alpha $ is a relative $r$-cycle by Lemma 62.6.2. Assume each $g_ i^*\alpha $ is a relative $r$-cycle. Let $g : S' \to S$ be a morphism where $S'$ is the spectrum of a discrete valuation ring. After replacing $S$ by $S'$, $X$ by $X' = X \times _ S S'$, and $\alpha $ by $\alpha ' = g^*\alpha $ and using that the base change of a h covering is a h covering (More on Flatness, Lemma 38.34.9) we reduce to the problem studied in the next paragraph.

Assume $S$ is the spectrum of a discrete valuation ring with closed point $0$ and generic point $\eta $. We have to show that $sp_{X/S}(\alpha _\eta ) = \alpha _0$. Since a h covering is a V covering (by definition), there is an $i$ and a specialization $s' \leadsto s$ of points of $S_ i$ with $g_ i(s') = \eta $ and $g_ i(s) = 0$, see Topologies, Lemma 34.10.13. By Properties, Lemma 28.5.10 we can find a morphism $h : S' \to S_ i$ from the spectrum $S'$ of a discrete valuation ring which maps the generic point $\eta '$ to $s'$ and maps the closed point $0'$ to $s$. Denote $\alpha ' = h^*g_ i^*\alpha $. By assumption we have $sp_{X'/S'}(\alpha '_{\eta '}) = \alpha '_{0'}$. Since $g = g_ i \circ h : S' \to S$ is the morphism of schemes induced by an extension of discrete valuation rings we conclude that $sp_{X/S}$ and $sp_{X'/S'}$ are compatible with base change maps on the fibres, see Lemma 62.4.3. We conclude that $sp_{X/S}(\alpha _\eta ) = \alpha _0$ because the base change map $Z_ r(X_0) \to Z_ r(X'_{0'})$ is injective as discussed in Section 62.3. $\square$

Proof. If $\alpha $ is a relative $r$-cycle, then each pull back $f_ i^*\alpha $ is a relative $r$-cycle by Lemma 62.6.2. Assume each $f_ i^*\alpha $ is a relative $r$-cycle. Let $g : S' \to S$ be a morphism where $S'$ is the spectrum of a discrete valuation ring. After replacing $S$ by $S'$, $X$ by $X' = X \times _ S S'$, and $\alpha $ by $\alpha ' = g^*\alpha $ we reduce to the problem studied in the next paragraph.

Assume $S$ is the spectrum of a discrete valuation ring with closed point $0$ and generic point $\eta $. We have to show that $sp_{X/S}(\alpha _\eta ) = \alpha _0$. Denote $f_{i, 0} : X_{i, 0} \to X_0$ the base change of $f_ i$ to the closed point of $S$. Similarly for $f_{i, \eta }$. Observe that

Namely, the first equality holds by Lemma 62.4.4 and the second by assumption. Since the family of maps $f_{i, 0}^* : Z_ r(X_0) \to Z_ r(X_{i, 0})$ is jointly injective (due to the fact that $f_{i, 0}$ is jointly surjective), we conclude what we want. $\square$

Proof. Since base change commutes with $i_*$ (Lemma 62.5.1) it suffices to prove the following: if $S$ is the spectrum of a discrete valuation ring with generic point $\eta $ and closed point $0$, then $sp_{X/S}(\alpha _\eta ) = \alpha _0$ if and only if $sp_{Y/S}(i_{\eta , *}\alpha _\eta ) = i_{0, *}\alpha _0$. This is true because $i_{0, *} : Z_ r(X_0) \to Z_ r(Y_0)$ is injective and because $i_{0, *}sp_{X/S}(\alpha _\eta ) = sp_{Y/S}(i_{\eta , *}\alpha _\eta )$ by Lemma 62.4.5. $\square$

Proof. The implication (1) $\Rightarrow $ (2) is immediate. Assume (2). For every $s \in S$ we can find an $\eta $ as in (2) which specializes to $s$. By Properties, Lemma 28.5.10 we can find a morphism $g : S' \to S$ from the spectrum $S'$ of a discrete valuation ring which maps the generic point $\eta '$ to $\eta $ and maps the closed point $0$ to $s$. Then $\alpha _ s$ and $\beta _ s$ are elements of $Z_ r(X_ s)$ which base change to the same element of $Z_ r(X_{0'})$, namely $sp_{X_{S'}/S'}(\alpha _{\eta '})$ where $\alpha _{\eta '}$ is the base change of $\alpha _\eta $. Since the base change map $Z_ r(X_ s) \to Z_ r(X_{0'})$ is injective as discussed in Section 62.3 we conclude $\alpha _ s = \beta _ s$. $\square$

Proof. By More on Flatness, Lemma 38.20.9 the hypothesis on $\mathcal{F}$ is preserved by any base change. Also, formation of $[\mathcal{F}/X/S]_ r$ is compatible with any base change by Lemma 62.5.3. Since the condition of being compatible with specializations is checked after base change to the spectrum of a discrete valuation ring, this reduces us to the case where $S$ is the spectrum of a valuation ring. In this case the set $U = \{ x \in X \mid \mathcal{F}\text{ flat at }x\text{ over }S\} $ is open in $X$ by More on Flatness, Lemma 38.13.11. Since the complement of $U$ in $X$ has fibres of dimension $< r$ over $S$ by assumption, we see that restriction along the inclusion $U \subset X$ induces an isomorphism on the groups of $r$-cycles on fibres after any base change, compatible with specialization maps and with formation of the relative cycle associated to $\mathcal{F}$. Thus it suffices to show compability with specializations for $[\mathcal{F}|_ U / U /S]_ r$. Since $\mathcal{F}|_ U$ is flat over $S$, this follows from Lemma 62.4.1 and the definitions. $\square$

Proof. The assumption means that $\mathcal{O}_ Z$ is flat over $S$ in dimensions $\geq r$. Thus applying Lemma 62.6.7 with $\mathcal{F} = (Z \to X)_*\mathcal{O}_ Z$ we conclude. $\square$

Proof. By Noetherian induction, we may assume the result holds for the pullback of $\alpha $ by any closed immersion $g : S' \to S$ which is not an isomorphism.

Let $S_1 \subset S$ be an irreducible component (viewed as an integral closed subscheme). Let $S_2 \subset S$ be the closure of the complement of $S'$ (viewed as a reduced closed subscheme). If $S_2 \not= \emptyset $, then the result holds for the pullback of $\alpha $ by $S_1 \to S$ and $S_2 \to S$. If $g_1 : S'_1 \to S_1$ and $g_2 : S'_2 \to S_2$ are the corresponding completely decomposed proper morphisms, then $S' = S'_1 \amalg S'_2 \to S$ is a completely decomposed proper morphism and we see the result holds for $S$¹ . Thus we may assume $S' \to S$ is bijective and we reduce to the case described in the next paragraph.

Assume $S$ is integral. Let $\eta \in S$ be the generic point and let $K = \kappa (\eta )$ be the function field of $S$. Then $\alpha _\eta $ is an $r$-cycle on $X_ K$. Write $\alpha _\eta = \sum n_ i[Y_ i]$. Taking the closure of $Y_ i$ we obtain integral closed subschemes $Z_ i \subset X$ whose base change to $\eta $ is $Y_ i$. By generic flatness (for example Morphisms, Proposition 29.27.1), we see that $Z_ i$ is flat over a nonempty open $U$ of $S$ for each $i$. Applying More on Flatness, Lemma 38.31.1 we can find a $U$-admissible blowing up $g : S' \to S$ such that the strict transform $Z'_ i \subset X_{S'}$ of $Z_ i$ is flat over $S'$. Then $\beta = \sum n_ i[Z'_ i/X_{S'}/S']_ r$ is in the image of (62.6.8.1) and $\beta = g^*\alpha $ by Lemma 62.6.6.

However, this does not finish the proof as $S' \to S$ may not be completely decomposed. This is easily fixed: denoting $T \subset S$ the complement of $U$ (viewed as a closed subscheme), by Noetherian induction we can find a completely decomposed proper morphism $T' \to T$ such that $(T' \to S)^*\alpha $ is in the image of (62.6.8.1). Then $S' \amalg T' \to S$ does the job. $\square$

Proof. This of course follows from Lemma 62.6.9 but we can also see it directly as follows. Say $\alpha $ is a relative $r$-cycle on $X/S$. Write $\alpha _\eta = \sum n_ i[Z_ i]$ (the sum is finite). Denote $\overline{Z}_ i \subset X$ the closure of $Z_ i$ as in Section 62.4. Then $\alpha = \sum n_ i[\overline{Z}_ i/X/S]$. $\square$

Proof. Let $x \in X$ with image $s \in S$. Since $f$ is smooth, there is a unique irreducible component $Z(x)$ of $X_ s$ which contains $x$. Then $\dim (Z(x)) = r$. Let $n_ x$ be the coefficient of $Z(x)$ in the cycle $\alpha _ s$. We will show the function $x \mapsto n_ x$ is locally constant on $X$.

Let $g : S' \to S$ be a morphism of locally Noetherian schemes. Let $X'$ be the base change of $X$ and let $\alpha ' = g^*\alpha $ be the base change of $\alpha $. Let $x' \in X'$ map to $s' \in S'$, $x \in X$, and $s \in S$. We claim $n_{x'} = n_ x$. Namely, since $Z(x)$ is smooth over $\kappa (s)$ we see that $Z(x) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} \mathop{\mathrm{Spec}}(\kappa (s'))$ is reduced. Since $Z(x')$ is an irreducible component of this scheme, we see that the coefficient $n_{x'}$ of $Z(x')$ in $\alpha '_{s'}$ is the same as the coefficient $n_ x$ of $Z(x)$ in $\alpha _ s$ by the definition of base change in Section 62.3 thereby proving the claim.

Since $X$ is locally Noetherian, to show that $x \mapsto n_ x$ is locally constant, it suffices to show: if $x' \leadsto x$ is a specialization in $X$, then $n_{x'} = n_ x$. Choose a morphism $S' \to X$ where $S'$ is the spectrum of a discrete valuation ring mapping the generic point $\eta $ to $x'$ and the closed point $0$ to $x$. See Properties, Lemma 28.5.10. Then the base change $X' \to S'$ of $f$ by $S' \to S$ has a section $\sigma : S' \to X'$ such that $\sigma (\eta ) \leadsto \sigma (0)$ is a specialization of points of $X'$ mapping to $x' \leadsto x$ in $X$. Thus we reduce to the claim in the next paragraph.

Let $S$ be the spectrum of a discrete valuation ring with generic point $\eta $ and closed point $0$ and we have a section $\sigma : S \to X$. Claim: $n_{\sigma (\eta )} = n_{\sigma (0)}$. By the discussion in More on Morphisms, Section 37.29 and especially More on Morphisms, Lemma 37.29.6 after replacing $X$ by an open subscheme, we may assume the fibres of $X \to S$ are connected. Since these fibres are smooth, they are irreducible. Then we see that $\alpha _\eta = n[X_\eta ]$ with $n = n_{\sigma (\eta )}$ and the relation $sp_{X/S}(\alpha _\eta ) = \alpha _0$ implies $\alpha _0 = n[X_0]$, i.e., $n_{\sigma (0)} = n$ as desired. $\square$

Proof. The question is local on $S$, thus we may assume $S$ is affine. Let $X = \bigcup U_ i$ be an affine open covering. Let $E_ i = \{ s \in S : \alpha _ s|_{U_{i, s}} = \beta _ s|_{U_{i, s}}\} $. Then $E = \bigcap E_ i$. Hence it suffices to prove the lemma for $U_ i \to S$ and the restriction of $\alpha $ and $\beta $ to $U_ i$. This reduces us to the case discussed in the next paragraph.

Assume $X$ and $S$ are quasi-compact. Set $\gamma = \alpha - \beta $. Then $E = \{ s \in S : \gamma _ s = 0\} $. By Lemma 62.6.8 there exists a jointly surjective finite family of proper morphisms $\{ g_ i : S_ i \to S\} $ such that $g_ i^*\gamma $ is in the image of (62.6.8.1). Observe that $E_ i = g_ i^{-1}(E)$ is the set of point $t \in S_ i$ such that $(g_ i^*\gamma )_ t = 0$. If $E_ i$ is closed for all $i$, then $E = \bigcup g_ i(E_ i)$ is closed as well. This reduces us to the case discussed in the next paragraph.

Assume $X$ and $S$ are quasi-compact and $\gamma = \sum n_ i[Z_ i/X/S]_ r$ for a finite number of closed subschemes $Z_ i \subset X$ flat and of relative dimension $\leq r$ over $S$. Set $X' = \bigcup Z_ i$ (scheme theoretic union). Then $i : X' \to X$ is a closed immersion and $X'$ has relative dimension $\leq r$ over $S$. Also $\gamma = i_*\gamma '$ where $\gamma ' = \sum n_ i[Z_ i/X'/S]_ r$. Since clearly $E = E' = \{ s \in S : \gamma '_ s = 0\} $ we reduce to the case discussed in the next paragraph.

Assume $X$ has relative dimension $\leq r$ over $S$. Let $s \in S$, $s \not\in E$. We will show that there exists an open neighbourhood $V \subset S$ of $s$ such that $E \cap V$ is empty. The assumption $s \not\in E$ means there exists an integral closed subscheme $Z \subset X_ s$ of dimension $r$ such that the coefficient $n$ of $[Z]$ in $\gamma _ s$ is nonzero. Let $x \in Z$ be the generic point. Since $\dim (Z) = r$ we see that $x$ is a generic point of an irreducible component (namely $Z$) of $X_ s$. Thus after replacing $X$ by an open neighbourhood of $x$, we may assume that $Z$ is the only irreducible component of $X_ s$. In particular, we have $\gamma _ s = n[Z]$.

At this point we apply More on Morphisms, Lemma 37.47.1 and we obtain a diagram

with all the properties listed there. Let $\gamma ' = g^*\gamma $ be the flat pullback. Note that $E \subset E' = \{ s \in S: \gamma '_ s = 0\} $ and that $s \not\in E'$ because the coefficient of $Z'$ in $\gamma '_ s$ is nonzero, where $Z' \subset X'_ s$ is the closure of $x'$. Similarly, set $\gamma '' = \pi _*\gamma '$. Then we have $E' \subset E'' = \{ s \in S: \gamma ''_ s = 0\} $ and $s \not\in E''$ because the coefficient of $Z''$ in $\gamma ''_ s$ is nonzero, where $Z'' \subset Y_ s$ is the closure of $y$. By Lemma 62.6.11 and openess of $Y \to S$ we see that an open neighbourhood of $s$ is disjoint from $E''$ and the proof is complete. $\square$

Proof. Suppose that $i \geq 0$ and $\alpha _ i, \beta _ i \in z(X_ i/S_ i, r)$ map to the same element of $z(X/S, r)$. Then $S \to S_ i$ maps into the closed subset $E \subset S_ i$ of Lemma 62.6.12. Hence for some $j \geq i$ the morphism $S_ j \to S_ i$ maps into $E$, see Limits, Lemma 32.4.10. It follows that the base change of $\alpha _ i$ and $\beta _ i$ to $S_ j$ agree. Thus the map is injective.

Let $\alpha \in z(X/S, r)$. Applying Lemma 62.6.9 a completely decomposed proper morphism $g : S' \to S$ such that $g^*\alpha $ is in the image of (62.6.8.1). Set $X' = S' \times _ S X$. We write $g^*\alpha = \sum n_ a [Z_ a/X'/S']_ r$ for some $Z_ a \subset X'$ closed subscheme flat and of relative dimension $\leq r$ over $S'$.

Now we bring the machinery of Limits, Section 32.10 ff to bear. We can find an $i \geq 0$ such that there exist

1. a completely decomposed proper morphism $g_ i : S'_ i \to S_ i$ whose base change to $S$ is $g : S' \to S$,

2. setting $X'_ i = S'_ i \times _{S_ i} X_ i$ closed subschemes $Z_{ai} \subset X'_ i$ flat and of relative dimension $\leq r$ over $S'_ i$ whose base change to $S'$ is $Z_ a$.

To do this one uses Limits, Lemmas 32.10.1, 32.8.5, 32.8.7, 32.13.1, and 32.18.1 and More on Morphisms, Lemma 37.78.5. Consider $\alpha '_ i = \sum n_ a [Z_{ai}/X'_ i/S'_ i]_ r \in z(X'_ i/S'_ i, r)$. The image of $\alpha '_ i$ in $z(X'/S', r)$ agrees with the base change $g^*\alpha $ by construction.

Set $S''_ i = S'_ i \times _{S_ i} S'_ i$ and $X''_ i = S''_ i \times _{S_ i} X_ i$ and set $S'' = S' \times _ S S'$ and $X'' = S'' \times _ S X$. We denote $\text{pr}_1, \text{pr}_2 : S'' \to S'$ and $\text{pr}_1, \text{pr}_2 : S''_ i \to S'_ i$ the projections. The two base changes $\text{pr}_1^*\alpha '_ i$ and $\text{pr}_1^*\alpha '_ i$ map to the same element of $z(X''/S'', r)$ because $\text{pr}_1^*g^*\alpha = \text{pr}_1^*g^*\alpha $. Hence after increasing $i$ we may assume that $\text{pr}_1^*\alpha '_ i = \text{pr}_1^*\alpha '_ i$ by the first paragraph of the proof. By Lemma 62.5.9 we obtain a unique family $\alpha _ i$ of $r$-cycles on fibres of $X_ i/S_ i$ with $g_ i^*\alpha _ i = \alpha '_ i$ (this uses that $S'_ i \to S_ i$ is completely decomposed). By Lemma 62.6.3 we see that $\alpha _ i \in z(X_ i/S_ i, r)$. The uniqueness in Lemma 62.5.9 implies that the image of $\alpha _ i$ in $z(X/S, r)$ is $\alpha $ and the proof is complete. $\square$

Proof. Since $i_ s : X_ s \to X'_ s$ is a thickening it is clear that $i_*$ induces a bijection between families of $r$-cycles on the fibres of $X/S$ and families of $r$-cycles on the fibres of $X'/S$. Also, given a family $\alpha $ of $r$-cycles on the fibres of $X/S$ $\alpha \in z(X/S, r) \Leftrightarrow i_*\alpha \in z(X'/S, r)$ by Lemma 62.6.5. The lemma follows. $\square$

Proof. Since $Z_ r(X_ s) \to Z_ r(U_ s)$ is a bijection by the dimension assumption, we see that restriction induces a bijection between families of $r$-cycles on the fibres of $X/S$ and families of $r$-cycles on the fibres of $U/S$. These restriction maps $Z_ r(X_ s) \to Z_ r(U_ s)$ are compatible with base change and with specializations, see Lemma 62.5.1 and 62.4.4. The lemma follows easily from this; details omitted. $\square$

Proof. By Lemma 62.5.10 we have a bijection between the group of families of $r$-cycles on fibres of $X/S$ and the group of families of $r$-cycles on fibres of $X'/S'$. Say $\alpha $ is a families of $r$-cycles on fibres of $X/S$ and $\alpha ' = g^*\alpha $ is the base change. If $R$ is a discrete valuation ring, then any morphism $h : \mathop{\mathrm{Spec}}(R) \to S$ factors as $g \circ h'$ for some unique morphism $h' : \mathop{\mathrm{Spec}}(R) \to S'$. Namely, the morphism $S' \times _ S \mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(R)$ is a univeral homomorphism inducing bijections on residue fields, and hence has a section (for example because $R$ is a seminormal ring, see Morphisms, Section 29.47). Thus the condition that $\alpha $ is compatible with specializations (i.e., is a relative $r$-cycle) is equivalent to the condition that $\alpha '$ is compatible with specializations. $\square$