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In the paper, Notes on etale cohomology of number fields, Mazur insists that for a finite etale morphism $\pi$, the Norm theorem holds (on. p.543, Remarks (e)).

More specifically, let $\pi : Y \rightarrow X$ be a finite morphism, where $Y=\text{Spec}(D_Y)$ and $X=\text{Spec} (D_X)$, and $D$'s are Dedekind domains. For a constructible sheaf $F$ on $Y$, there are the Norm maps $$ N_r : \text{Ext}^r_Y (F, \mathbb{G}_m) \rightarrow \text{Ext}^r_X (\pi_* F, \mathbb{G}_m) $$ (on. p.542 of the same paper). If $\pi$ is etale, then $N_r$ is an isomorphism for any $r$. Mazur commented that a beautiful morphism $F \rightarrow \pi^*\pi_*F$ provides an inverse to $N_r$.

Can someone explain this in details?

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One has an isomorphism $$N_r: {\rm{Ext}}^r_Y(F, \pi^{\ast}G) \rightarrow {\rm{Ext}}^r_X(\pi_{\ast}F, G)$$ for finite etale $\pi:Y \rightarrow X$ between arbitrary schemes and any abelian etale sheaves $F$ on $Y$ and $G$ on $X$, without needing to appeal to an explicit procedure for an inverse map (though the argument below adapts to show that it is given by composing $\pi$-pullback with a canonical map $F \rightarrow \pi^{\ast}\pi_{\ast}F$ to which Mazur refers).

The idea is to reduce the problem to a sheaf-theoretic isomorphism, so one can then work locally on the base to pass to the case of a split covering that is easy. (This uses that we got away from the original setting in which $Y$ is required to be irreducible.)

Let's first recall how $N_r$ is defined (in the spirit of coherent duality). Since $\mathbf{R}{\rm{Hom}}_Y = \mathbf{R}\Gamma_Y \circ \mathbf{R}\mathscr{H}om_Y$ and $\Gamma_Y = \Gamma_X \circ \pi_*$ with $\pi_*$ exact (as $\pi$ is finite), we define $N_r$ via passage to $r$th homology on the effect of applying $\mathbf{R}\Gamma_X$ to the natural sheaf map $$\pi_{\ast}\mathbf{R}\mathscr{H}om_Y(F, \pi^{\ast}(\cdot)) \rightarrow \mathbf{R}\mathscr{H}om_X(\pi_{\ast}F, \pi_{\ast}\pi^{\ast}(\cdot)) \rightarrow \mathbf{R}\mathscr{H}om_X(\pi_{\ast}F, \cdot)$$ using the sheaf-theoretic trace for finite etale maps (or even just quasi-finite etale separated maps) at the final step.

Thus, to show $N_r$ is an isomorphism for finite etale $\pi$ it suffices to show that the above displayed map is an isomorphism. But this task is of etale-local nature on $X$, and since $\pi$ is finite etale we can pass to an etale cover on $X$ to arrive at the case that $\pi$ is a split covering; i.e., $Y$ is a finite disjoint union of copies $X_1, \dots, X_n$ of $X$. Now $F$ amounts to a collection of sheaves $F_1, \dots, F_n$ on the $X_i$'s and the map in question is the natural composite map $$\oplus_i \mathbf{R}\mathscr{H}om_X(F_i, G) \rightarrow \mathbf{R}\mathscr{H}om_X(\oplus_i F_i, \oplus_i G) \rightarrow \mathbf{R}\mathscr{H}om_X(\oplus_i F_i, G)$$ (using the addition map $\oplus_i G \rightarrow G$ at the final step) for any abelian etale sheaves $F_1, \dots, F_n, G$ on $X$. But this is obviously an isomorphism.

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  • $\begingroup$ So the map $F\to\pi^*\pi_*F$ corresponds to $F_i\mapsto \oplus_j F_j$; is that correct? $\endgroup$ – Jason Starr Apr 11 '17 at 12:07
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    $\begingroup$ @JasonStarr: Yes. For any quasi-compact separated etale map of schemes $f$ one has a natural map ${\rm{id}} \rightarrow f^! f_!$, and for finite etale $f$ we naturally have $f_! = f_*$ and $f^! = f^*$. For $f$ a split finite etale cover this map becomes exactly what you have described (but that isn't the definition). $\endgroup$ – nfdc23 Apr 11 '17 at 12:23

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