The Stacks project

Proof. In the statement $f(Z) \subset X$ is the scheme theoretic image of $f$ and it is also the reduced induced algebraic space structure on the closed subset $f(|Z|) \subset X$, see Morphisms of Spaces, Lemma 67.16.4. Then $Z$ and $f(Z)$ are reduced, proper (hence decent) algebraic spaces over $k$, whence integral (Definition 72.4.1). The left hand side is computed using the coefficient of $n_1 \ldots n_ d$ in the function

The equality follows from Lemma 72.17.3 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 72.18.1 and the result is true. Assume $\dim (f(Z)) = d$. Then by dimension theory (Lemma 72.15.2) we find that the equivalent conditions (1) – (5) of Lemma 72.5.1 hold. Thus $\deg (Z \to f(Z))$ is well defined. By the already used Lemma 72.5.1 we find $f : Z \to f(Z)$ is finite over a nonempty open $V$ of $f(Z)$; after possibly shrinking $V$ we may assume $V$ is a scheme. Let $\xi \in V$ be the generic point. Thus $\deg (f : Z \to f(Z))$ the length of the stalk of $f_*\mathcal{O}_ Z$ at $\xi $ over $\mathcal{O}_{X, \xi }$ and the stalk of $R^ if_*\mathcal{O}_ X$ at $\xi $ is zero for $i > 0$ (for example by Cohomology of Spaces, Lemma 69.4.1). 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 72.18.2. The desired result follows. $\square$