I heard that if $\mathbf F_q$ is a finite field, $X, Y$ are birational smooth proper variety over $\mathbf F_q$, then $\#(X(\mathbf F_q)) \equiv \#(Y(\mathbf F_q)) \pmod q$, and I heard that the proof uses crystalline cohomology. Anyone could explain the proof, or is there a reference, thanks.
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2$\begingroup$ This is not true. The projective plane and the plane blown up at a point are a counterexample. They probably meant the number of rational points mod $q$ or something like that. $\endgroup$– Will SawinOct 19, 2021 at 2:17

$\begingroup$ @Sawin I'm sorry, fixed it. $\endgroup$– Aoi KoshigayaOct 19, 2021 at 2:36

2$\begingroup$ See lovelylittlelemmas.rjprojects.net/… $\endgroup$– Damian RösslerOct 19, 2021 at 11:58

$\begingroup$ @Rössler Thanks very much. By the way, the reference Sur le groupe fondamental d’une variété unirationelle mentioned in the link, I couldn't find this article on the internet, is there an online version of it. $\endgroup$– Aoi KoshigayaOct 19, 2021 at 12:41

1$\begingroup$ @越谷あおい drive.google.com/file/d/1Z1sBvkv7ZRihZhuznbR3vWol2Nuoee1z/… although I believe you can find old comptes rendus papers officially somewhere $\endgroup$– FrankOct 19, 2021 at 15:46
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