## 68.4 Finite morphisms

Here are some results which hold for all abelian sheaves (in particular also quasi-coherent modules). We warn the reader that these lemmas do not hold for finite morphisms of schemes and the Zariski topology.

Lemma 68.4.1. Let $S$ be a scheme. Let $f : X \to Y$ be an integral (for example finite) morphism of algebraic spaces. Then $f_* : \textit{Ab}(X_{\acute{e}tale}) \to \textit{Ab}(Y_{\acute{e}tale})$ is an exact functor and $R^ pf_* = 0$ for $p > 0$.

Proof. By Properties of Spaces, Lemma 65.18.12 we may compute the higher direct images on an étale cover of $Y$. Hence we may assume $Y$ is a scheme. This implies that $X$ is a scheme (Morphisms of Spaces, Lemma 66.45.3). In this case we may apply Étale Cohomology, Lemma 59.43.5. For the finite case the reader may wish to consult the less technical Étale Cohomology, Proposition 59.55.2. $\square$

Lemma 68.4.2. Let $S$ be a scheme. Let $f : X \to Y$ be a finite morphism of algebraic spaces over $S$. Let $\overline{y}$ be a geometric point of $Y$ with lifts $\overline{x}_1, \ldots , \overline{x}_ n$ in $X$. Then

$(f_*\mathcal{F})_{\overline{y}} = \prod \nolimits _{i = 1, \ldots , n} \mathcal{F}_{\overline{x}_ i}$

for any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$.

Proof. Choose an étale neighbourhood $(V, \overline{v})$ of $\overline{y}$. Then the stalk $(f_*\mathcal{F})_{\overline{y}}$ is the stalk of $f_*\mathcal{F}|_ V$ at $\overline{v}$. By Properties of Spaces, Lemma 65.18.12 we may replace $Y$ by $V$ and $X$ by $X \times _ Y V$. Then $Z \to X$ is a finite morphism of schemes and the result is Étale Cohomology, Proposition 59.55.2. $\square$

Lemma 68.4.3. Let $S$ be a scheme. Let $\pi : X \to Y$ be a finite morphism of algebraic spaces over $S$. Let $\mathcal{A}$ be a sheaf of rings on $X_{\acute{e}tale}$. Let $\mathcal{B}$ be a sheaf of rings on $Y_{\acute{e}tale}$. Let $\varphi : \mathcal{B} \to \pi _*\mathcal{A}$ be a homomorphism of sheaves of rings so that we obtain a morphism of ringed topoi

$f = (\pi , \varphi ) : (\mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}), \mathcal{A}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (Y_{\acute{e}tale}), \mathcal{B}).$

For a sheaf of $\mathcal{A}$-modules $\mathcal{F}$ and a sheaf of $\mathcal{B}$-modules $\mathcal{G}$ the canonical map

$\mathcal{G} \otimes _\mathcal {B} f_*\mathcal{F} \longrightarrow f_*(f^*\mathcal{G} \otimes _\mathcal {A} \mathcal{F}).$

is an isomorphism.

Proof. The map is the map adjoint to the map

$f^*\mathcal{G} \otimes _\mathcal {A} f^* f_*\mathcal{F} = f^*(\mathcal{G} \otimes _\mathcal {B} f_*\mathcal{F}) \longrightarrow f^*\mathcal{G} \otimes _\mathcal {A} \mathcal{F}$

coming from $\text{id} : f^*\mathcal{G} \to f^*\mathcal{G}$ and the adjunction map $f^* f_*\mathcal{F} \to \mathcal{F}$. To see this map is an isomorphism, we may check on stalks (Properties of Spaces, Theorem 65.19.12). Let $\overline{y}$ be a geometric point of $Y$ and let $\overline{x}_1, \ldots , \overline{x}_ n$ be the geometric points of $X$ lying over $\overline{y}$. Working out what our maps does on stalks, we see that we have to show

$\mathcal{G}_{\overline{y}} \otimes _{\mathcal{B}_{\overline{y}}} \left( \bigoplus \nolimits _{i = 1, \ldots , n} \mathcal{F}_{\overline{x}_ i} \right) = \bigoplus \nolimits _{i = 1, \ldots , n} (\mathcal{G}_{\overline{y}} \otimes _{\mathcal{B}_{\overline{x}}} \mathcal{A}_{\overline{x}_ i}) \otimes _{\mathcal{A}_{\overline{x}_ i}} \mathcal{F}_{\overline{x}_ i}$

which holds true. Here we have used that taking tensor products commutes with taking stalks, the behaviour of stalks under pullback Properties of Spaces, Lemma 65.19.9, and the behaviour of stalks under pushforward along a closed immersion Lemma 68.4.2. $\square$

We end this section with an insanely general projection formula for finite morphisms.

Lemma 68.4.4. With $S$, $X$, $Y$, $\pi$, $\mathcal{A}$, $\mathcal{B}$, $\varphi$, and $f$ as in Lemma 68.4.3 we have

$K \otimes _\mathcal {B}^\mathbf {L} Rf_*M = Rf_*(Lf^*K \otimes _\mathcal {A}^\mathbf {L} M)$

in $D(\mathcal{B})$ for any $K \in D(\mathcal{B})$ and $M \in D(\mathcal{A})$.

Proof. Since $f_*$ is exact (Lemma 68.4.1) the functor $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a complex $\mathcal{K}^\bullet$ of $\mathcal{B}$-modules representing $K$ which is K-flat with flat terms, see Cohomology on Sites, Lemma 21.17.11. Then $f^*\mathcal{K}^\bullet$ is K-flat with flat terms, see Cohomology on Sites, Lemma 21.18.1. Choose any complex $\mathcal{M}^\bullet$ of $\mathcal{A}$-modules representing $M$. Then we have to show

$\text{Tot}(\mathcal{K}^\bullet \otimes _\mathcal {B} f_*\mathcal{M}^\bullet ) = f_*\text{Tot}(f^*\mathcal{K}^\bullet \otimes _\mathcal {A} \mathcal{M}^\bullet )$

because by our choices these complexes represent the right and left hand side of the formula in the lemma. Since $f_*$ commutes with direct sums (for example by the description of the stalks in Lemma 68.4.2), this reduces to the equalities

$\mathcal{K}^ n \otimes _\mathcal {B} f_*\mathcal{M}^ m = f_*(f^*\mathcal{K}^ n \otimes _\mathcal {A} \mathcal{M}^ m)$

which are true by Lemma 68.4.3. $\square$

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