Lemma 81.11.2. In Situation 81.2.1 let $X, Y/B$ be good. Let $f : X \to Y$ be a finite locally free morphism of degree $d$ (see Morphisms of Spaces, Definition 66.46.2). Then $f$ is both proper and flat of relative dimension $0$, and

\[ f_*f^*\alpha = d\alpha \]

for every $\alpha \in Z_ k(Y)$.

**Proof.**
A finite locally free morphism is flat and finite by Morphisms of Spaces, Lemma 66.46.6, and a finite morphism is proper by Morphisms of Spaces, Lemma 66.45.9. We omit showing that a finite morphism has relative dimension $0$. Thus the formula makes sense. To prove it, let $Z \subset Y$ be an integral closed subscheme of $\delta $-dimension $k$. It suffices to prove the formula for $\alpha = [Z]$. Since the base change of a finite locally free morphism is finite locally free (Morphisms of Spaces, Lemma 66.46.5) we see that $f_*f^*\mathcal{O}_ Z$ is a finite locally free sheaf of rank $d$ on $Z$. Thus clearly $f_*f^*\mathcal{O}_ Z$ has length $d$ at the generic point of $Z$. Hence

\[ f_*f^*[Z] = f_*f^*[\mathcal{O}_ Z]_ k = [f_*f^*\mathcal{O}_ Z]_ k = d[Z] \]

where we have used Lemmas 81.10.5 and 81.8.3.
$\square$

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