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 $S$.

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 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. $\square$

Comment #2450 by John Smith on

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

Comment #2492 by on

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 $R \setminus \mathfrak p$.

Comment #3797 by Vincenzo Zaccaro on

How can the coefficient of the term $Z$ be $\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$? Why is it not equal to $\deg(Z/f(Z))\cdot\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$?

Comment #3798 by Vincenzo Zaccaro on

How can the coefficient of the term $Z$ be $\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$? Why is it not equal to $\deg(Z/f(Z))\cdot\text{length}_{R_{\mathfrak p}}(M_{\mathfrak p})$?

Comment #3799 by Vincenzo Zaccaro on

Oh sorry for the double comment! Anyway I meant $\deg(X/f(X))$.

Comment #3917 by on

Can you explain more precisely what the problem is? I think everything is correct. Can you work out what you are saying in an example? (If the error is as bad as you are suggesting then almost any example will be a counter example.)

There are also:

• 2 comment(s) on Section 42.12: Proper pushforward

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).