Let $X$ be a smooth projective variety over a finite field $k$, and $F$ the geometric Frobenius on $H^*_{et}(X_{k^{sep}}, \mathbf{Z}_{\ell})$.
Is $F$ only rationally bijective, or integrally bijective too?
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2$\begingroup$ You are right about the degree but the Frobenious is sort of degree a power of p, and we assume of course that $\ell \ne p$. $\endgroup$– S. carmeliCommented Mar 31, 2018 at 19:26
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3$\begingroup$ The "smooth projective" condition and constant sheaves and condition $\ell \ne p$ are all red herrings. On ${\rm{H}}^i(X_{k_s}, \mathscr{F}_{k_s})$ for any $k$-scheme $X$ and any abelian etale sheaf $\mathscr{F}$ on $X$ the effect of pullback of any element of ${\rm{Gal}}(k_s/k)$ is obviously an automorphism, and for the Galois element $\phi: t \mapsto t^{1/q}$ and $\mathscr{F}$ constant that coincides with the effect of pullback along the scalar extension of the $q$-Frobenius endomorphism of $X$: see section 1.5.1 in math.ru.nl/~bmoonen/Seminars/EtCohConrad.pdf $\endgroup$– nfdc23Commented Mar 31, 2018 at 22:14
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1$\begingroup$ @nfdc23 OK. However, let's assume $X$ smooth projective, is defined over $\mathbf{Q}$, and carries a lift $\phi$ of the absolute Frobenius of a mod $p$ fiber, for some $p$ of good reduction. Base change $X$ to $\mathbf{C}$ and get $H^*(X_{\mathbf{C}}^{an},\mathbf{Z})$, which after scalar extension to $\mathbf{Z}_{\ell}$ agrees with $H^*(X_{\overline{\mathbf{Q}}},\mathbf{Z}_{\ell})$. On the left, the effect of $\phi\otimes\mathbf{C}$ is carried on the inverse of the Galois element $t\mapsto t^p$ on $\mathbf{Z}_{\ell}$-etale cohomology. However, $\phi\otimes\mathbf{C}$ cannot act as automorphism $\endgroup$– user120812Commented Apr 1, 2018 at 0:49
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1$\begingroup$ But it does after scalar extension from $\mathbf{Z}$ to $\mathbf{Z}_{\ell}$, and my question is, really: this is due to the fact that $\phi\otimes\mathbf{C}$ has $p$-power degree on $X_{\mathbf{C}}^{an}$, and $p$ is a unit in $\mathbf{Z}_{\ell}$, as long as $\ell\neq p$. Right? $\endgroup$– user120812Commented Apr 1, 2018 at 0:51
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2$\begingroup$ Yes, arguing via a finite flat map having $p$-power degree is a way to explain it, but this feels backwards (one would never prove the fact over finite fields using the strong hypothesis of such char. 0 lifts) since one can prove the statement directly on the cohomology of the geometric mod-$p$ fiber (and then use smooth and proper base change). I think a more appropriate way to think about the point you're making is that it explains why the result in characteristic $p$ is consistent with a separate conclusion one can obtain by analytic means in char. 0. Anyway, you seem to understand it fine. $\endgroup$– nfdc23Commented Apr 1, 2018 at 2:27
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