Lemma 57.86.1. 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 $S_ L$.

## 57.86 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 10.153.11. Thus the lemma in this case follows immediately from Lemma 57.85.3. In the general case, let $K'$ and $L'$ be the perfect closures (Algebra, Definition 10.44.5) 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 10.45.7). 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 57.45.4. This reduces the lemma to the case of the field extension $L'/K'$ which is separable (by definition of perfect fields, see Algebra, Definition 10.44.1).
$\square$

Lemma 57.86.2. 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 10.45.7). 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 57.45.4. 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 10.153.11. We conclude our lemma is a special case of Theorem 57.85.2 as reformulated in Lemma 57.85.3.

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 57.86.3. With $f : X \to S$ and $n$ as in Remark 57.84.1 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 57.84.7. 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 57.86.2 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. $\square$

## 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)