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.
$\square$

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