Lemma 33.45.7. Let $k$ be a field. Let $f : Y \to X$ be a morphism of proper schemes over $k$. Let $Z \subset Y$ be an integral closed subscheme of dimension $d$ and let $\mathcal{L}_1, \ldots , \mathcal{L}_ d$ be invertible $\mathcal{O}_ X$-modules. Then
where $\deg (Z \to f(Z))$ is as in Morphisms, Definition 29.51.8 or $0$ if $\dim (f(Z)) < d$.
Proof. The left hand side is computed using the coefficient of $n_1 \ldots n_ d$ in the function
The equality follows from Lemma 33.33.5 and the projection formula (Cohomology, Lemma 20.54.2). If $f(Z)$ has dimension $< d$, then the right hand side is a polynomial of total degree $< d$ by Lemma 33.45.1 and the result is true. Assume $\dim (f(Z)) = d$. Let $\xi \in f(Z)$ be the generic point. By dimension theory (see Lemmas 33.20.3 and 33.20.4) the generic point of $Z$ is the unique point of $Z$ mapping to $\xi $. Then $f : Z \to f(Z)$ is finite over a nonempty open of $f(Z)$, see Morphisms, Lemma 29.51.1. Thus $\deg (f : Z \to f(Z))$ is defined and in fact it is equal to the length of the stalk of $f_*\mathcal{O}_ Z$ at $\xi $ over $\mathcal{O}_{X, \xi }$. Moreover, the stalk of $R^ if_*\mathcal{O}_ X$ at $\xi $ is zero for $i > 0$ because we just saw that $f|_ Z$ is finite in a neighbourhood of $\xi $ (so that Cohomology of Schemes, Lemma 30.9.9 gives the vanishing). Thus the terms $\chi (X, R^ if_*\mathcal{O}_ Z \otimes \mathcal{L}_1^{\otimes n_1} \otimes \ldots \otimes \mathcal{L}_ d^{\otimes n_ d})$ with $i > 0$ have total degree $< d$ and
modulo a polynomial of total degree $< d$ by Lemma 33.45.2. The desired result follows. $\square$
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