Lemma 58.67.2. 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 58.66.12 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 $\mathcal{O}_{X, \bar\eta } = \kappa (\eta )^{sep}$ and thus

since the corresponding Galois group is trivial. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)