Lemma 59.90.1. Let $L/K$ be an extension of fields. Let $g : T \to S$ be a quasi-compact and quasi-separated morphism of schemes over $K$. Denote $g_ L : T_ L \to S_ L$ the base change of $g$ to $\mathop{\mathrm{Spec}}(L)$. Let $E \in D^+(T_{\acute{e}tale})$ have cohomology sheaves whose stalks are torsion of orders invertible in $K$. Let $E_ L$ be the pullback of $E$ to $(T_ L)_{\acute{e}tale}$. Then $Rg_{L, *}E_ L$ is the pullback of $Rg_*E$ to $S_ L$.

## 59.90 Applications of smooth base change

In this section we discuss some more or less immediate consequences of the smooth base change theorem.

**Proof.**
If $L/K$ is separable, then $L$ is a filtered colimit of smooth $K$-algebras, see Algebra, Lemma 10.158.11. Thus the lemma in this case follows immediately from Lemma 59.89.3. In the general case, let $K'$ and $L'$ be the perfect closures (Algebra, Definition 10.45.5) of $K$ and $L$. Then $\mathop{\mathrm{Spec}}(K') \to \mathop{\mathrm{Spec}}(K)$ and $\mathop{\mathrm{Spec}}(L') \to \mathop{\mathrm{Spec}}(L)$ are universal homeomorphisms as $K'/K$ and $L'/L$ are purely inseparable (see Algebra, Lemma 10.46.7). Thus we have $(T_{K'})_{\acute{e}tale}= T_{\acute{e}tale}$, $(S_{K'})_{\acute{e}tale}= S_{\acute{e}tale}$, $(T_{L'})_{\acute{e}tale}= (T_ L){\acute{e}tale}$, and $(S_{L'})_{\acute{e}tale}= (S_ L)_{\acute{e}tale}$ by the topological invariance of étale cohomology, see Proposition 59.45.4. This reduces the lemma to the case of the field extension $L'/K'$ which is separable (by definition of perfect fields, see Algebra, Definition 10.45.1).
$\square$

Lemma 59.90.2. Let $K/k$ be an extension of separably closed fields. Let $X$ be a quasi-compact and quasi-separated scheme over $k$. Let $E \in D^+(X_{\acute{e}tale})$ have cohomology sheaves whose stalks are torsion of orders invertible in $k$. Then

the maps $H^ q_{\acute{e}tale}(X, E) \to H^ q_{\acute{e}tale}(X_ K, E|_{X_ K})$ are isomorphisms, and

$E \to R(X_ K \to X)_*E|_{X_ K}$ is an isomorphism.

**Proof.**
Proof of (1). First let $\overline{k}$ and $\overline{K}$ be the algebraic closures of $k$ and $K$. The morphisms $\mathop{\mathrm{Spec}}(\overline{k}) \to \mathop{\mathrm{Spec}}(k)$ and $\mathop{\mathrm{Spec}}(\overline{K}) \to \mathop{\mathrm{Spec}}(K)$ are universal homeomorphisms as $\overline{k}/k$ and $\overline{K}/K$ are purely inseparable (see Algebra, Lemma 10.46.7). Thus $H^ q_{\acute{e}tale}(X, \mathcal{F}) = H^ q_{\acute{e}tale}(X_{\overline{k}}, \mathcal{F}_{X_{\overline{k}}})$ by the topological invariance of étale cohomology, see Proposition 59.45.4. Similarly for $X_ K$ and $X_{\overline{K}}$. Thus we may assume $k$ and $K$ are algebraically closed. In this case $K$ is a limit of smooth $k$-algebras, see Algebra, Lemma 10.158.11. We conclude our lemma is a special case of Theorem 59.89.2 as reformulated in Lemma 59.89.3.

Proof of (2). For any quasi-compact and quasi-separated $U$ in $X_{\acute{e}tale}$ the above shows that the restriction of the map $E \to R(X_ K \to X)_*E|_{X_ K}$ determines an isomorphism on cohomology. Since every object of $X_{\acute{e}tale}$ has an étale covering by such $U$ this proves the desired statement. $\square$

Lemma 59.90.3. With $f : X \to S$ and $n$ as in Remark 59.88.1 assume $n$ is invertible on $S$ and that for some $q \geq 1$ we have that $BC(f, n, q - 1)$ is true, but $BC(f, n, q)$ is not. Then there exist a commutative diagram

with both squares cartesian, where $S'$ is affine, integral, and normal with algebraically closed function field $K$ and there exists an integer $d | n$ such that $R^ qh_*(\mathbf{Z}/d\mathbf{Z})$ is nonzero.

**Proof.**
First choose a diagram and $\mathcal{F}$ as in Lemma 59.88.7. We may and do assume $S'$ is affine (this is obvious, but see proof of the lemma in case of doubt). Let $K'$ be the function field of $S'$ and let $Y' = X' \times _{S'} \mathop{\mathrm{Spec}}(K')$ to get the diagram

By Lemma 59.90.2 the total direct image $R(Y \to Y')_*\mathbf{Z}/d\mathbf{Z}$ is isomorphic to $\mathbf{Z}/d\mathbf{Z}$ in $D(Y'_{\acute{e}tale})$; here we use that $n$ is invertible on $S$. Thus $Rh'_*\mathbf{Z}/d\mathbf{Z} = Rh_*\mathbf{Z}/d\mathbf{Z}$ by the relative Leray spectral sequence. This finishes the proof. $\square$

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