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Tag 02R6

Chapter 41: Chow Homology and Chern Classes > Section 41.13: Proper pushforward

Lemma 41.13.3. Let $(S, \delta)$ be as in Situation 41.8.1. Let $f : X \to Y$ be a proper morphism of schemes which are locally of finite type over $S$.

  1. Let $Z \subset X$ be a closed subscheme with $\dim_\delta(Z) \leq k$. Then $$ f_*[Z]_k = [f_*{\mathcal O}_Z]_k. $$
  2. Let $\mathcal{F}$ be a coherent sheaf on $X$ such that $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$. Then $$ f_*[\mathcal{F}]_k = [f_*{\mathcal F}]_k. $$

Note that the statement makes sense since $f_*\mathcal{F}$ and $f_*\mathcal{O}_Z$ are coherent $\mathcal{O}_Y$-modules by Cohomology of Schemes, Proposition 29.19.1.

Proof. Part (1) follows from (2) and Lemma 41.11.3. Let $\mathcal{F}$ be a coherent sheaf on $X$. Assume that $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$. By Cohomology of Schemes, Lemma 29.9.7 there exists a closed subscheme $i : Z \to X$ and a coherent $\mathcal{O}_Z$-module $\mathcal{G}$ such that $i_*\mathcal{G} \cong \mathcal{F}$ and such that the support of $\mathcal{F}$ is $Z$. Let $Z' \subset Y$ be the scheme theoretic image of $f|_Z : Z \to Y$. Consider the commutative diagram of schemes $$ \xymatrix{ Z \ar[r]_i \ar[d]_{f|_Z} & X \ar[d]^f \\ Z' \ar[r]^{i'} & Y } $$ We have $f_*\mathcal{F} = f_*i_*\mathcal{G} = i'_*(f|_Z)_*\mathcal{G}$ by going around the diagram in two ways. Suppose we know the result holds for closed immersions and for $f|_Z$. Then we see that $$ f_*[\mathcal{F}]_k = f_*i_*[\mathcal{G}]_k = (i')_*(f|_Z)_*[\mathcal{G}]_k = (i')_*[(f|_Z)_*\mathcal{G}]_k = [(i')_*(f|_Z)_*\mathcal{G}]_k = [f_*\mathcal{F}]_k $$ as desired. The case of a closed immersion is straightforward (omitted). Note that $f|_Z : Z \to Z'$ is a dominant morphism (see Morphisms, Lemma 28.6.3). Thus we have reduced to the case where $\dim_\delta(X) \leq k$ and $f : X \to Y$ is proper and dominant.

Assume $\dim_\delta(X) \leq k$ and $f : X \to Y$ is proper and dominant. Since $f$ is dominant, for every irreducible component $Z \subset Y$ with generic point $\eta$ there exists a point $\xi \in X$ such that $f(\xi) = \eta$. Hence $\delta(\eta) \leq \delta(\xi) \leq k$. Thus we see that in the expressions $$ f_*[\mathcal{F}]_k = \sum n_Z[Z], \quad \text{and} \quad [f_*\mathcal{F}]_k = \sum m_Z[Z]. $$ whenever $n_Z \not = 0$, or $m_Z \not = 0$ the integral closed subscheme $Z$ is actually an irreducible component of $Y$ of $\delta$-dimension $k$. Pick such an integral closed subscheme $Z \subset Y$ and denote $\eta$ its generic point. Note that for any $\xi \in X$ with $f(\xi) = \eta$ we have $\delta(\xi) \geq k$ and hence $\xi$ is a generic point of an irreducible component of $X$ of $\delta$-dimension $k$ as well (see Lemma 41.10.1). Since $f$ is quasi-compact and $X$ is locally Noetherian, there can be only finitely many of these and hence $f^{-1}(\{\eta\})$ is finite. By Morphisms, Lemma 28.48.1 there exists an open neighbourhood $\eta \in V \subset Y$ such that $f^{-1}(V) \to V$ is finite. Replacing $Y$ by $V$ and $X$ by $f^{-1}(V)$ we reduce to the case where $Y$ is affine, and $f$ is finite.

Write $Y = \mathop{\rm Spec}(R)$ and $X = \mathop{\rm Spec}(A)$ (possible as a finite morphism is affine). Then $R$ and $A$ are Noetherian rings and $A$ is finite over $R$. Moreover $\mathcal{F} = \widetilde{M}$ for some finite $A$-module $M$. Note that $f_*\mathcal{F}$ corresponds to $M$ viewed as an $R$-module. Let $\mathfrak p \subset R$ be the minimal prime corresponding to $\eta \in Y$. The coefficient of $Z$ in $[f_*\mathcal{F}]_k$ is clearly $\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$. Let $\mathfrak q_i$, $i = 1, \ldots, t$ be the primes of $A$ lying over $\mathfrak p$. Then $A_{\mathfrak p} = \prod A_{\mathfrak q_i}$ since $A_{\mathfrak p}$ is an Artinian ring being finite over the dimension zero local Noetherian ring $R_{\mathfrak p}$. Clearly the coefficient of $Z$ in $f_*[\mathcal{F}]_k$ is $$ \sum\nolimits_{i = 1, \ldots, t} [\kappa(\mathfrak q_i) : \kappa(\mathfrak p)] \text{length}_{A_{\mathfrak q_i}}(M_{\mathfrak q_i}) $$ Hence the desired equality follows from Algebra, Lemma 10.51.12. $\square$

    The code snippet corresponding to this tag is a part of the file chow.tex and is located in lines 3539–3560 (see updates for more information).

    \begin{lemma}
    \label{lemma-cycle-push-sheaf}
    Let $(S, \delta)$ be as in Situation \ref{situation-setup}.
    Let $f : X \to Y$ be a proper morphism of schemes which are
    locally of finite type over $S$.
    \begin{enumerate}
    \item Let $Z \subset X$ be a closed subscheme with $\dim_\delta(Z) \leq k$.
    Then
    $$
    f_*[Z]_k = [f_*{\mathcal O}_Z]_k.
    $$
    \item Let $\mathcal{F}$ be a coherent sheaf on $X$ such that
    $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$. Then
    $$
    f_*[\mathcal{F}]_k = [f_*{\mathcal F}]_k.
    $$
    \end{enumerate}
    Note that the statement makes sense since $f_*\mathcal{F}$ and
    $f_*\mathcal{O}_Z$ are coherent $\mathcal{O}_Y$-modules by
    Cohomology of Schemes, Proposition
    \ref{coherent-proposition-proper-pushforward-coherent}.
    \end{lemma}
    
    \begin{proof}
    Part (1) follows from (2) and Lemma \ref{lemma-cycle-closed-coherent}.
    Let $\mathcal{F}$ be a coherent sheaf on $X$.
    Assume that $\dim_\delta(\text{Supp}(\mathcal{F})) \leq k$.
    By Cohomology of Schemes, Lemma \ref{coherent-lemma-coherent-support-closed}
    there exists a closed subscheme $i : Z \to X$ and a coherent
    $\mathcal{O}_Z$-module $\mathcal{G}$ such that
    $i_*\mathcal{G} \cong \mathcal{F}$ and such that the support
    of $\mathcal{F}$ is $Z$. Let $Z' \subset Y$ be the scheme theoretic image
    of $f|_Z : Z \to Y$. Consider the commutative diagram of schemes
    $$
    \xymatrix{
    Z \ar[r]_i \ar[d]_{f|_Z} &
    X \ar[d]^f \\
    Z' \ar[r]^{i'} & Y
    }
    $$
    We have $f_*\mathcal{F} = f_*i_*\mathcal{G} = i'_*(f|_Z)_*\mathcal{G}$
    by going around the diagram in two ways. Suppose we know the result holds
    for closed immersions and for $f|_Z$. Then we see that
    $$
    f_*[\mathcal{F}]_k = f_*i_*[\mathcal{G}]_k
    = (i')_*(f|_Z)_*[\mathcal{G}]_k =
    (i')_*[(f|_Z)_*\mathcal{G}]_k =
    [(i')_*(f|_Z)_*\mathcal{G}]_k = [f_*\mathcal{F}]_k
    $$
    as desired. The case of a closed immersion is straightforward (omitted).
    Note that $f|_Z : Z \to Z'$ is a dominant morphism (see
    Morphisms, Lemma \ref{morphisms-lemma-quasi-compact-scheme-theoretic-image}).
    Thus we have reduced to the case where
    $\dim_\delta(X) \leq k$ and $f : X \to Y$ is proper and dominant.
    
    \medskip\noindent
    Assume $\dim_\delta(X) \leq k$ and $f : X \to Y$ is proper and dominant.
    Since $f$ is dominant, for every irreducible component $Z \subset Y$
    with generic point $\eta$ there exists a point $\xi \in X$ such
    that $f(\xi) = \eta$. Hence $\delta(\eta) \leq \delta(\xi) \leq k$.
    Thus we see that in the expressions
    $$
    f_*[\mathcal{F}]_k = \sum n_Z[Z],
    \quad
    \text{and}
    \quad
    [f_*\mathcal{F}]_k = \sum m_Z[Z].
    $$
    whenever $n_Z \not = 0$, or $m_Z \not = 0$ the integral closed
    subscheme $Z$ is actually an irreducible component of $Y$ of
    $\delta$-dimension $k$. Pick such an integral closed subscheme
    $Z \subset Y$ and denote $\eta$ its generic point. Note that for
    any $\xi \in X$ with $f(\xi) = \eta$ we have $\delta(\xi) \geq k$
    and hence $\xi$ is a generic point of an irreducible component
    of $X$ of $\delta$-dimension $k$ as well
    (see Lemma \ref{lemma-multiplicity-finite}). Since $f$ is quasi-compact
    and $X$ is locally Noetherian, there can be only finitely many of
    these and hence $f^{-1}(\{\eta\})$ is finite.
    By Morphisms, Lemma \ref{morphisms-lemma-generically-finite} there exists
    an open neighbourhood $\eta \in V \subset Y$ such that $f^{-1}(V) \to V$
    is finite. Replacing $Y$ by $V$ and $X$ by $f^{-1}(V)$ we reduce to the
    case where $Y$ is affine, and $f$ is finite.
    
    \medskip\noindent
    Write $Y = \Spec(R)$ and $X = \Spec(A)$ (possible as
    a finite morphism is affine).
    Then $R$ and $A$ are Noetherian rings and $A$ is finite over $R$.
    Moreover $\mathcal{F} = \widetilde{M}$ for some finite $A$-module
    $M$. Note that $f_*\mathcal{F}$ corresponds to $M$ viewed as an $R$-module.
    Let $\mathfrak p \subset R$ be the minimal prime corresponding
    to $\eta \in Y$. The coefficient of $Z$ in $[f_*\mathcal{F}]_k$
    is clearly $\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$.
    Let $\mathfrak q_i$, $i = 1, \ldots, t$ be the primes of $A$
    lying over $\mathfrak p$. Then $A_{\mathfrak p} = \prod A_{\mathfrak q_i}$
    since $A_{\mathfrak p}$ is an Artinian ring being finite over the
    dimension zero local Noetherian ring $R_{\mathfrak p}$.
    Clearly the coefficient of $Z$ in $f_*[\mathcal{F}]_k$ is
    $$
    \sum\nolimits_{i = 1, \ldots, t}
    [\kappa(\mathfrak q_i) : \kappa(\mathfrak p)]
    \text{length}_{A_{\mathfrak q_i}}(M_{\mathfrak q_i})
    $$
    Hence the desired equality follows from
    Algebra, Lemma \ref{algebra-lemma-pushdown-module}.
    \end{proof}

    Comments (2)

    Comment #2450 by John Smith on March 10, 2017 a 10:19 pm UTC

    How can $A_{\mathfrak{p}}$ be a ring when $\mathfrak{p}$ is not a prime ideal in $A$?

    Comment #2492 by Johan (site) on April 13, 2017 a 11:06 pm UTC

    This is an often used abuse of notation. If $R \to A$ is a ring map and $\mathfrak p \subset R$ is a prime ideal, then $A_\mathfrak p$ is the localization of $A$ at $\mathfrak p$ as an $R$-module. Thus $A_\mathfrak p = (R \setminus \mathfrak p)^{-1}A$. However, this is also a ring because this is equal to $S^{-1}A$ where $S \subset A$ is the image of $R \setminus \mathfrak p$.

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