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