The Stacks project

Lemma 59.97.6. Let $K$ be a field. For any commutative diagram

\[ \xymatrix{ X \ar[d] & X' \ar[l] \ar[d]_{f'} & Y \ar[l]^ h \ar[d]^ e \\ \mathop{\mathrm{Spec}}(K) & S' \ar[l] & T \ar[l]_ g } \]

of schemes over $K$ with $X' = X \times _{\mathop{\mathrm{Spec}}(K)} S'$ and $Y = X' \times _{S'} T$ and $g$ quasi-compact and quasi-separated, and every abelian sheaf $\mathcal{F}$ on $T_{\acute{e}tale}$ whose stalks are torsion of orders invertible in $K$ the base change map

\[ (f')^{-1}Rg_*\mathcal{F} \longrightarrow Rh_*e^{-1}\mathcal{F} \]

is an isomorphism.

Proof. The question is local on $X$, hence we may assume $X$ is affine. By Limits, Lemma 32.7.2 we can write $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as a cofiltered limit with affine transition morphisms of schemes $X_ i$ of finite type over $K$. Denote $X'_ i = X_ i \times _{\mathop{\mathrm{Spec}}(K)} S'$ and $Y_ i = X'_ i \times _{S'} T$. By Lemma 59.86.3 it suffices to prove the statement for the squares with corners $X_ i, Y_ i, S_ i, T_ i$. Thus we may assume $X$ is of finite type over $K$. Similarly, we may write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}[n]$ where the colimit is over the multiplicative system of integers invertible in $K$. The same lemma used above reduces us to the case where $\mathcal{F}$ is a sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules for some $n$ invertible in $K$.

We may replace $K$ by its algebraic closure $\overline{K}$. Namely, formation of direct image commutes with base change to $\overline{K}$ according to Lemma 59.90.1 (works for both $g$ and $h$). And it suffices to prove the agreement after restriction to $X'_{\overline{K}}$. Next, we may replace $X$ by its reduction as we have the topological invariance of étale cohomology, see Proposition 59.45.4. After this replacement the morphism $X \to \mathop{\mathrm{Spec}}(K)$ is flat, finite presentation, with geometrically reduced fibres and the same is true for any base change, in particular for $X' \to S'$. Hence $(f')^{-1}g_*\mathcal{F} \to Rh_*e^{-1}\mathcal{F}$ is an isomorphism by Lemma 59.87.2.

At this point we may apply Lemma 59.90.3 to see that it suffices to prove: given a commutative diagram

\[ \xymatrix{ X \ar[d]_ f & X' \ar[d] \ar[l] & Y \ar[l]^ h \ar[d] \\ \mathop{\mathrm{Spec}}(K) & S' \ar[l] & \mathop{\mathrm{Spec}}(L) \ar[l] } \]

with both squares cartesian, where $S'$ is affine, integral, and normal with algebraically closed function field $K$, then $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ is zero for $q > 0$ and $d | n$. Observe that this vanishing is equivalent to the statement that

\[ (f')^{-1}R^ q(\mathop{\mathrm{Spec}}(L) \to S')_*\mathbf{Z}/d\mathbf{Z} \longrightarrow R^ qh_*\mathbf{Z}/d\mathbf{Z} \]

is an isomorphism, because the left hand side is zero for example by Lemma 59.80.5.

Write $S' = \mathop{\mathrm{Spec}}(B)$ so that $L$ is the fraction field of $B$. Write $B = \bigcup _{i \in I} B_ i$ as the union of its finite type $K$-subalgebras $B_ i$. Let $J$ be the set of pairs $(i, g)$ where $i \in I$ and $g \in B_ i$ nonzero with ordering $(i', g') \geq (i, g)$ if and only if $i' \geq i$ and $g$ maps to an invertible element of $(B_{i'})_{g'}$. Then $L = \mathop{\mathrm{colim}}\nolimits _{(i, g) \in J} (B_ i)_ g$. For $j = (i, g) \in J$ set $S_ j = \mathop{\mathrm{Spec}}(B_ i)$ and $U_ j = \mathop{\mathrm{Spec}}((B_ i)_ g)$. Then

\[ \vcenter { \xymatrix{ X' \ar[d] & Y \ar[l]^ h \ar[d] \\ S' & \mathop{\mathrm{Spec}}(L) \ar[l] } } \quad \text{is the colimit of}\quad \vcenter { \xymatrix{ X \times _ K S_ j \ar[d] & X \times _ K U_ j \ar[l]^{h_ j} \ar[d] \\ S_ j & U_ j \ar[l] } } \]

Thus we may apply Lemma 59.86.3 to see that it suffices to prove base change holds in the diagrams on the right which is what we proved in Lemma 59.97.5. $\square$

Comments (0)

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F1I. Beware of the difference between the letter 'O' and the digit '0'.