Lemma 37.75.1. Given a cartesian square
with $\pi $ locally quasi-finite with finite fibres and a function $w : U \to \mathbf{Z}$ we have $(\int _\pi w) \circ g = \int _{\pi '} (w \circ h)$.
Lemma 37.75.1. Given a cartesian square
with $\pi $ locally quasi-finite with finite fibres and a function $w : U \to \mathbf{Z}$ we have $(\int _\pi w) \circ g = \int _{\pi '} (w \circ h)$.
Proof. This follows immediately from the second description of $\int _\pi w$ given above. To prove it from the definition, you use that if $E/F$ is a finite extension of fields and $F'/F$ is another field extension, then writing $(E \otimes _ F F')_{red} = \prod E'_ i$ as a product of fields finite over $F'$, we have
To prove this equality pick an algebraically closed field extension $\Omega /F'$ and observe that
where we have used Fields, Lemma 9.14.8. $\square$
Comments (0)