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Let $f: X \to Y$ be a finite morphism of varieties with $Y$ smooth. Is there a projection formula for $f$ and $H^{i}_{et}(-,\mathbf{Z}(1)) = H^i(-,\mathbf{G}_m)$?

Background: I want to show that for a relative curve $\pi: C \to X$ ($C$, $X$, $\pi$ smooth) with a quasi-section $Y \to C$ closed immersion with $Y \to X$ finite of degree $d$, the kernel of $\pi^*: H^i(X,\mathbf{G}_m) \to H^i(C,\mathbf{G}_m)$ is annihilated by $d$.

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(As you have also stated the application you have in mind I assume that $X$ has everywhere the same dimension as $Y$ with no embedded components.)

We have that $H^i_{et}(X,\mathbf{G}_m) = H^i_{et}(Y,f_\ast\mathbf{G}_m)$ as $f$ is finite (this is because $f_\ast$ is exact, see Cor. II:3.5 of Milne: Étale cohomology). If $f$ is also flat we have a norm map $f_\ast\mathbf{G}_m \to \mathbf{G}_m$ whose composite with the inclusion $\mathbf{G}_m \to f_\ast\mathbf{G}_m$ is the $d$'th power which gives what you want. In the general case you still have a norm map by first noting that $f$ is flat in codimension $1$ (this comes from the condition I added) and you can take the norm there which then will land in $\mathbf{G}_m$ as $Y$ is normal.

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"We have that $H^i{et}(X,\mathbf{G}_m) = H^i{et}(Y,f_*\mathbf{G}_m)$ as $f$ is finite": I don't understand this. Why do the higher direct images vanish? I only know this for $f$ affine and the sheaf quasi-coherent. Doesn't $Y = \mathrm{Spec}\mathbf{Z}_p$, $X =$ non-normal finite extension of $Y$ give a counterexample for $R^1\f_*\mathbf{G}_m$? –  Timo Keller Apr 23 '10 at 15:30
@norondion: I have added elaboration in answer. –  Torsten Ekedahl Apr 23 '10 at 20:59
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