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Aoi Koshigaya
Aoi Koshigaya
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Number of rational points over finite fields mod $q$ is birational invariant

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)) = \#(Y(\mathbf F_q))$$\#(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.

Number of rational points over finite fields is birational invariant

I heard that if $\mathbf F_q$ is a finite field, $X, Y$ are birational smooth proper variety over $\mathbf F_q$, then $\#(X(\mathbf F_q)) = \#(Y(\mathbf F_q))$, and I heard that the proof uses crystalline cohomology. Anyone could explain the proof, or is there a reference, thanks.

Number of rational points over finite fields mod $q$ is birational invariant

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.

Source Link
Aoi Koshigaya
Aoi Koshigaya
  • 988
  • 4
  • 9

Number of rational points over finite fields is birational invariant

I heard that if $\mathbf F_q$ is a finite field, $X, Y$ are birational smooth proper variety over $\mathbf F_q$, then $\#(X(\mathbf F_q)) = \#(Y(\mathbf F_q))$, and I heard that the proof uses crystalline cohomology. Anyone could explain the proof, or is there a reference, thanks.