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. $\square$

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## Comments (2)

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