Lemma 59.85.2. Let $A$ be a henselian local ring. Let $X = \mathbf{P}^1_ A$. Let $X_0 \subset X$ be the closed fibre. Let $\ell$ be a prime number. Let $\mathcal{I}$ be an injective sheaf of $\mathbf{Z}/\ell \mathbf{Z}$-modules on $X_{\acute{e}tale}$. Then $H^ q_{\acute{e}tale}(X_0, \mathcal{I}|_{X_0}) = 0$ for $q > 0$.

Proof. Observe that $X$ is a separated scheme which can be covered by $2$ affine opens. Hence for $q > 1$ this follows from Gabber's affine variant of the proper base change theorem, see Lemma 59.82.8. Thus we may assume $q = 1$. Let $\xi \in H^1_{\acute{e}tale}(X_0, \mathcal{I}|_{X_0})$. Goal: show that $\xi$ is $0$. By Lemmas 59.73.2 and 59.51.4 we can find a map $\mathcal{F} \to \mathcal{I}$ with $\mathcal{F}$ a constructible sheaf of $\mathbf{Z}/\ell \mathbf{Z}$-modules and $\xi$ coming from an element $\zeta$ of $H^1_{\acute{e}tale}(X_0, \mathcal{F}|_{X_0})$. Suppose we have an injective map $\mathcal{F} \to \mathcal{F}'$ of sheaves of $\mathbf{Z}/\ell \mathbf{Z}$-modules on $X_{\acute{e}tale}$. Since $\mathcal{I}$ is injective we can extend the given map $\mathcal{F} \to \mathcal{I}$ to a map $\mathcal{F}' \to \mathcal{I}$. In this situation we may replace $\mathcal{F}$ by $\mathcal{F}'$ and $\zeta$ by the image of $\zeta$ in $H^1_{\acute{e}tale}(X_0, \mathcal{F}'|_{X_0})$. Also, if $\mathcal{F} = \mathcal{F}_1 \oplus \mathcal{F}_2$ is a direct sum, then we may replace $\mathcal{F}$ by $\mathcal{F}_ i$ and $\zeta$ by the image of $\zeta$ in $H^1_{\acute{e}tale}(X_0, \mathcal{F}_ i|_{X_0})$.

By Lemma 59.74.4 and the remarks above we may assume $\mathcal{F}$ is of the form $f_*\underline{M}$ where $M$ is a finite $\mathbf{Z}/\ell \mathbf{Z}$-module and $f : Y \to X$ is a finite morphism of finite presentation (such sheaves are still constructible by Lemma 59.73.9 but we won't need this). Since formation of $f_*$ commutes with any base change (Lemma 59.55.3) we see that the restriction of $f_*\underline{M}$ to $X_0$ is equal to the pushforward of $\underline{M}$ via the induced morphism $Y_0 \to X_0$ of special fibres. By the Leray spectral sequence (Proposition 59.54.2) and vanishing of higher direct images (Proposition 59.55.2), we find

$H^1_{\acute{e}tale}(X_0, f_*\underline{M}|_{X_0}) = H^1_{\acute{e}tale}(Y_0, \underline{M}).$

Since $Y \to \mathop{\mathrm{Spec}}(A)$ is proper we can use Lemma 59.85.1 to see that the $H^1_{\acute{e}tale}(Y_0, \underline{M})$ is equal to $H^1_{\acute{e}tale}(Y, \underline{M})$. Thus we see that our cohomology class $\zeta$ lifts to a cohomology class

$\tilde\zeta \in H^1_{\acute{e}tale}(Y, \underline{M}) = H^1_{\acute{e}tale}(X, f_*\underline{M})$

However, $\tilde\zeta$ maps to zero in $H^1_{\acute{e}tale}(X, \mathcal{I})$ as $\mathcal{I}$ is injective and by commutativity of

$\xymatrix{ H^1_{\acute{e}tale}(X, f_*\underline{M}) \ar[r] \ar[d] & H^1_{\acute{e}tale}(X, \mathcal{I}) \ar[d] \\ H^1_{\acute{e}tale}(X_0, (f_*\underline{M})|_{X_0}) \ar[r] & H^1_{\acute{e}tale}(X_0, \mathcal{I}|_{X_0}) }$

we conclude that the image $\xi$ of $\zeta$ is zero as well. $\square$

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).