Lemma 42.12.4. Let $(S, \delta )$ be as in Situation 42.7.1. Let $f : X \to Y$ be a proper morphism of schemes which are locally of finite type over $S$.

Let $Z \subset X$ be a closed subscheme with $\dim _\delta (Z) \leq k$. Then

\[ f_*[Z]_ k = [f_*{\mathcal O}_ Z]_ k. \]

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 30.19.1.

**Proof.**
Part (1) follows from (2) and Lemma 42.10.3. Let $\mathcal{F}$ be a coherent sheaf on $X$. Assume that $\dim _\delta (\text{Supp}(\mathcal{F})) \leq k$. By Cohomology of Schemes, Lemma 30.9.7 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 29.6.3). 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 42.9.1). 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 29.51.1 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 $Y = \mathop{\mathrm{Spec}}(R)$ and $X = \mathop{\mathrm{Spec}}(A)$ (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 10.52.12.
$\square$

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