The Stacks project

Proof. Let $W \subset X$ be an integral closed subscheme of dimension $k$. Consider $W' = f(W) \subset Y$ and $W'' = g(f(W)) \subset Z$. Since $f$, $g$ are proper we see that $W'$ (resp. $W''$) is an integral closed subscheme of $Y$ (resp. $Z$). We have to show that $g_*(f_*[W]) = (g \circ f)_*[W]$. If $\dim _\delta (W'') < k$, then both sides are zero. If $\dim _\delta (W'') = k$, then we see the induced morphisms

both satisfy the hypotheses of Lemma 42.11.1. Hence

Then we can apply Morphisms, Lemma 29.51.9 to conclude. $\square$

Proof. First assume $X$ is quasi-compact. Then $Z_ k(X)$ is a free $\mathbf{Z}$-module with basis given by the elements $[Z]$ where $Z \subset X$ is integral closed of $\delta$-dimension $k$. The groups $Z_ k(X_1)$, $Z_ k(X_2)$, $Z_ k(X_1 \cap X_2)$ are free on the subset of these $Z$ such that $Z \subset X_1$, $Z \subset X_2$, $Z \subset X_1 \cap X_2$. This immediately proves the lemma in this case. The general case is similar and the proof is omitted. $\square$

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

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

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

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

Hence the desired equality follows from Algebra, Lemma 10.52.12. $\square$