Lemma 59.68.2. For any q \geq 1, R^ q j_*\mathbf{G}_{m, \eta } = 0.
Proof. We need to show that (R^ q j_*\mathbf{G}_{m, \eta })_{\bar x} = 0 for every geometric point \bar x of X.
Assume that \bar x lies over a closed point x of X. Let \mathop{\mathrm{Spec}}(A) be an affine open neighbourhood of x in X, and K the fraction field of A. Then
The ring \mathcal{O}^{sh}_{X, \bar x} \otimes _ A K is a localization of the discrete valuation ring \mathcal{O}^{sh}_{X, \bar x}, so it is either \mathcal{O}^{sh}_{X, \bar x} again, or its fraction field K^{sh}_{\bar x}. But since some local uniformizer gets inverted, it must be the latter. Hence
Now recall that \mathcal{O}^{sh}_{X, \bar x} = \mathop{\mathrm{colim}}\nolimits _{(U, \bar u) \to \bar x} \mathcal{O}(U) = \mathop{\mathrm{colim}}\nolimits _{A \subset B} B where A \to B is étale, hence K^{sh}_{\bar x} is an algebraic extension of K = k(X), and we may apply Lemma 59.67.12 to get the vanishing.
Assume that \bar x = \bar\eta lies over the generic point \eta of X (in fact, this case is superfluous). Then \mathcal{O}^{sh}_{X, \bar\eta } = \kappa (\eta )^{sep} and thus
since the corresponding Galois group is trivial. \square
Comments (2)
Comment #8758 by Nico on
Comment #9321 by Stacks project on