Question on Hartshorne exercise II 5.17(e)

Question

I'm trying to solve exercise II 5.17(e) of Hartshorne:

Let $f:X\to Y$ be an affine morphism between schemes (i.e. preimage of every open affine subscheme of $Y$ is still affine), and let $\mathcal{A}=f_* \mathcal{O}_X$. Show that $f_*$ induces an equivalence of categories from the category of quasi-coherent $\mathcal{O}_X$-modules to the category of quasi-coherent $\mathcal{A}$-modules (i.e. quasi-coherent $\mathcal{O}_Y$-modules having a structure of $\mathcal{A}$-module).

In order to prove this question, it suffices to construct a quasi-coherent $\mathcal{O}_X$-module $\widetilde{\mathcal{M}}$ for any quasi-coherent $\mathcal{A}$-module $\mathcal{M}$, and show the functors $f_*$ and $\tilde{}$ are inverse to each other.

I am trying to construct $\widetilde{\mathcal{M}}$ by gluing sheaves: Since $\mathcal{M}$ is a quasi-coherent $\mathcal{A}$-module, $\mathcal{M}(U)$ has an $\mathcal{A}(U)=\mathcal{O}_X(f^{-1}(U))$- module structure. Then we have a sheaf $\widetilde{\mathcal{M}(U)}$ on affine open subset $f^{-1}(U)$. To glue up all these $\widetilde{\mathcal{M}(U)}$, we only need to check that, when $U'\subset U$ are open affine subsets, $\widetilde{\mathcal{M}(U)}|_{f^{-1}(U')}\simeq \widetilde{\mathcal{M}(U')}$. I am stuck here, and my attempt is as follows:

Let $U=\operatorname{Spec}B,U'=\operatorname{Spec}B',f^{-1}(U)=\operatorname{Spec}A,f^{-1}(U')=\operatorname{Spec}A'$.

$$\require{AMScd} \begin{CD} f^{-1}(U)=\operatorname{Spec}A @>{f_U}>> \operatorname{Spec}B=U\\ @AAA @AAA \\ f^{-1}(U')=\operatorname{Spec}A' @>{f_{U'}}>> \operatorname{Spec}B'=U' \end{CD}$$

$$\require{AMScd} \begin{CD} B=\mathcal{O}_Y(U) @>>> \mathcal{O}_X(f^{-1}(U))=A\\ @VVV @VVV \\ B'=\mathcal{O}_Y(U') @>>> \mathcal{O}_X(f^{-1}(U'))=A' \end{CD}$$

Since $\mathcal{M}$ is a quasi-coherent $\mathcal{A}$-module, $\mathcal{M}(U)$ has a $\mathcal{O}_Y(U)=B$-module structure via ring homomorphism $B=\mathcal{O}_Y(U)\to \mathcal{O}_X(f^{-1}(U))=A$. Then $\mathcal{M}$ is a quasi-coherent $\mathcal{O}_Y$-module implies that $\mathcal{M}|_U\simeq {f_U}_*(\widetilde{\mathcal{M}(U)})\simeq \widetilde{_B\mathcal{M}(U)}$, where $f_U:f^{-1}(U)=\operatorname{Spec}A\to\operatorname{Spec}B=U$. Similarly, we have $\mathcal{M}|_{U'}\simeq \widetilde{_{B'}\mathcal{M}(U')}$. Since $\mathcal{M}$ is quasi-coherent $\mathcal{O}_Y$-module, $(\mathcal{M}|_U)|_{U'}\simeq \widetilde{B'{\otimes{_B}_B}{\mathcal{M}(U)}}\simeq \mathcal{M}|_{U'} \simeq \widetilde{_{B'}\mathcal{M}(U')}$, which implies that $B'\otimes{_B} _B{\mathcal{M}(U)}\simeq {_{B'}\mathcal{M}(U')}$.

Similarly, $\widetilde{\mathcal{M}(U)}|_{f^{-1}(U')}\simeq \widetilde{\mathcal{M}(U')}$ is equivalent to $A'\otimes_A \mathcal{M}(U)\simeq \mathcal{M}(U')$, but how can we get this from $B'\otimes{_B} _B{\mathcal{M}(U)}\simeq {_{B'}\mathcal{M}(U')}$?

Edit

It seems that the most important fact in my proof is that, ${f_{U'}}_*(\widetilde{\mathcal{M}(U')})\simeq {f_U}_*(\widetilde{\mathcal{M}(U)})|_{U'}\simeq {f_{U'}}_*(\widetilde{\mathcal{M}(U)}|_{f^{-1}(U')})$. But how can we get $\widetilde{\mathcal{M}(U)}|_{f^{-1}(U')}\simeq \widetilde{\mathcal{M}(U')}$?

Answer 1

I'm not 100% sure I understand everything you wrote, but here is a way to simplify the situation:

To show that the sheaves glue, you actually only need to consider distinguished open sets $D_U(g)\subseteq U$ (for some $g\in\mathcal{O}_X(U)$). This is because for open affines $U,V\subseteq Y$, the affine opens $W\subseteq U\cap V$ which are distinguished both in $U$ and $V$ form a basis of the topology of $U\cap V$.

Now if you have a distinguished open $D_U(g)\subseteq U$, then $f^{-1}D_{U}(g)=D_{f^{-1}U}(f^{\#}(U)(g))$ and thus $$ \widetilde{\mathcal{M}(U)}(f^{-1}D_{U}(g))\cong\mathcal{M}(U)_{f^{\#}(U)(g)}. $$ But then as the $\mathcal{A}$-module structure is compatible with the $\mathcal{O}_Y$-module structure via $f^{\#}$, the localisation of $\mathcal{M}(U)$ at $f^{\#}(U)(g)$ as an $\mathcal{A}(U)$-module is the same as the localisation of $\mathcal{M}(U)$ at $g$ as an $\mathcal{O}_Y(U)$-module. But by quasi-coherence we have $\mathcal{M}(U)_g\cong \mathcal{M}(D_U(g))$, so we are done.

From there you can conclude the glueing, and you can also conclude the case where $U'\subseteq U$ is an arbitrary affine open. I don't know if there is a more straightforward way to do this.

Edit: Because it was wished by the OP, even more precisely:

Notice that $\mathcal{M}(U)_g$, which is a priori only an $\mathcal{O}_Y(U)_g$-module, naturally has the structure of an $\mathcal{A}(U)_{h}$-module, where $h:=f^{\#}(U)(g)$ for brevity. Indeed, we can define $$ \frac{a}{h^n}\cdot\frac{s}{g^m}:=\frac{as}{g^{m+n}}, $$ and verify that this endows $\mathcal{M}(U)_g$ with an $\mathcal{A}(U)_{h}$-module structure. It is then straightforward to see that $\mathcal{M}(U)_h\cong\mathcal{M}(U)_g$ as $\mathcal{A}(U)_h$-modules, and also $\mathcal{M}(U)_g\cong\mathcal{M}(D_U(g))$ as $\mathcal{A}(U)_h\cong\mathcal{A}(D_U(g))$-modules.

Edit 2:

To show that $\mathcal{M}(U)_g\to\mathcal{M}(D_U(g))$ is an $\mathcal{A}(U)_g\to\mathcal{M}(D_U(g))$-module homomorphism, denote by $\rho:\mathcal{M}(U)\to\mathcal{M}(D_U(g))$ the restriction map from $U$ to $D_U(g)$ (and by $\rho^\mathcal{A}$ resp. $\rho^Y$ the corresponding restriction maps of $\mathcal{A}$ resp. $\mathcal{O}_Y$). By definition of $\mathcal{M}$ being an $\mathcal{A}$-module, this is compatible with the action of $\mathcal{A}(U)$ on $\mathcal{M}(U)$ and $\mathcal{M}(D_U(g))$. The mentioned isomorphism is given by $$ \rho^{\mathcal{M}}_g:\mathcal{M}(U)_g\to\mathcal{M}(D_U(g))\\ \frac{s}{g^m}\mapsto \rho^{Y}(g)^{-m}\rho^{\mathcal{M}}(s). $$ Similarly, the isomorphism $\mathcal{A}(U)_h\to\mathcal{A}(D_U(g))$ is given by $$ \rho^{\mathcal{A}}_h:\mathcal{A}(U)_h\to\mathcal{A}(D_U(g))\\ \frac{s}{h^n}\mapsto \rho^{Y}(g)^{-n}\rho^{\mathcal{A}}(s). $$ Then we may verify that $$ \rho^{\mathcal{M}}_g\left(\frac{a}{h^n}\cdot\frac{s}{g^m}\right)=\rho^{\mathcal{M}}_g\left(\frac{as}{g^{m+n}}\right)=\rho^Y(g)^{-(m+n)}\rho^\mathcal{M}(as)=\rho^Y(g)^{-(m+n)}\rho^\mathcal{A}(a)\rho^{\mathcal{M}}(s)=\rho_h^{\mathcal{A}}(a/h^n)\rho_g^{\mathcal{M}}(s/g^m). $$ But all this blabla hides what is actually going on: you have to crucially use that $\mathcal{M}$ is an $\mathcal{A}$-module, so all the restriction maps are compatible with $\mathcal{A}$. Localisation doesn't change anything about this compatibility, because if muliplication is compatible, then division is as well, as soon as it is well defined.

Answer 2

$\def\Hom{\operatorname{Hom}} \def\Spec{\operatorname{Spec}}$Some months ago I sent the following proof of Tag 01SB to Professor A.J. de Jong so maybe they add it to the Stacks Project. In the meantime before it (might) get accepted, I will post it here in case it's useful to anyone.

Lemma. Let $S$ be a scheme and let $\mathcal{A}$ be a quasi-coherent $\mathcal{O}_S$-algebra. An $\mathcal{A}$-module is quasi-coherent as an $\mathcal{O}_S$-module if and only if it is quasi-coherent as an $\mathcal{A}$-module.

Proof. Let $\mathcal{F}$ be an $\mathcal{A}$-module. Assume first $\mathcal{A}$-quasi-coherence, and choose an open $U\subset S$ and an exact sequence $$ 0\to \bigoplus_J\mathcal{A}|_U\to \bigoplus_I\mathcal{A}|_U\to \mathcal{F}|_U $$ of $\mathcal{A}|_U$-modules. Then this is also an exact sequence of $\mathcal{O}_U$-modules. Hence, $\mathcal{F}|_U$ is the cokernel of a morphism of quasi-coherent $\mathcal{O}_U$-modules on a scheme and, thus, $\mathcal{O}_U$-quasi-coherent.

Conversely, assume $\mathcal{F}$ is $\mathcal{O}_X$-quasi-coherent. Pick an open affine $\Spec R=U\subset S$. We have isomorphisms of $\mathcal{O}_U$-modules $\mathcal{A}|_U\cong\widetilde{A}$ and $\mathcal{F}|_U\cong\widetilde{M}$, for some $R$-algebra $A$ and some $R$-module $M$. Since $\mathcal{F}|_U$ is also an $\mathcal{A}|_U$-module, the second isomorphism induces an $\mathcal{A}|_U$-module structure on $\widetilde{M}$ such that $\mathcal{F}|_U\cong\widetilde{M}$ promotes to an isomorphism of $\mathcal{A}|_U$-modules. In particular, $M$ is an $A$-module, and we can choose an exact sequence $$ 0\to \bigoplus_J A\to \bigoplus_I A\to M $$ of $A$-modules. Since localization is exact, for every distinguished open $D(f)\subset U$, $f\in R$, the last sequence induces an exact sequence $$ 0\to\bigoplus_J A_f\to\bigoplus_I A_f\to M_f $$ of $A_f$-modules. Hence we obtain an exact sequence $$ 0\to \bigoplus_J \widetilde{A}\to \bigoplus_I \widetilde{A}\to \widetilde{M} $$ of $\widetilde{A}$-modules. This means that $\mathcal{F}$ is $\mathcal{A}$-quasi-coherent. $\square$

Lemma. Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{A} = f_*\mathcal{O}_X$. The functor $\mathcal{F} \mapsto f_*\mathcal{F}$ induces an equivalence of categories $$ \left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{O}_X\text{-modules} \end{matrix} \right\} \longrightarrow \left\{ \begin{matrix} \text{category of quasi-coherent}\\ \mathcal{A}\text{-modules} \end{matrix} \right\}. $$

Proof. In the proof of Lemma 01S8 it was argued that affine morphisms are quasi-compact and quasi-separated. Hence, $f_*\mathcal{F}$ is a quasi-coherent $\mathcal{O}_S$-module by 01LC (in particular, $f_*\mathcal{O}_X$ is a quasi-coherent $\mathcal{O}_S$-algebra). By previous Lemma, $f_*\mathcal{F}$ is $\mathcal{A}$-quasi-coherent.

We see the fully faithfulness of the functor from the statement. That is, we verify that $\Hom_{\mathcal{O}_X}(\mathcal{F},\mathcal{G})\to \Hom_{f_*\mathcal{O}_X}(f_*\mathcal{F},f_*\mathcal{G})$ is bijective. It suffices to see that the morphism of sheaves on $S$ \begin{equation} \label{map-affine-equivalence-modules}\tag{1} \Hom_{\mathcal{O}_{f^{-1}(-)}} (\mathcal{F}|_{f^{-1}(-)}, \mathcal{G}|_{f^{-1}(-)})\to \mathcal{H}om_{f_*\mathcal{O}_X} (f_*\mathcal{F}, f_*\mathcal{G}) \end{equation} is an isomorphism. In turn, it suffices to see that $$ \Hom_{\mathcal{O}_{f^{-1}(U)}} (\mathcal{F}|_{f^{-1}(U)}, \mathcal{G}|_{f^{-1}(U)})\to \Hom_{f_*\mathcal{O}_X|_U} (f_*\mathcal{F}|_U, f_*\mathcal{G}|_U) $$ is bijective for $U\subset S$ open affine. Since $f^{-1}(U)$ is affine, we can assume $S$ is affine, $X=\Spec A$ and show that $\Hom_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})\to \Hom_{f_*\mathcal{O}_X}(f_*\widetilde{M},f_*\widetilde{N})$ is bijective, where $M,N$ are $A$-modules. The inverse is defined by considering a $f_*\mathcal{O}_X$-linear map $f_*\widetilde{M}\to f_*\widetilde{N}$, then taking the $A$-linear map on global sections $M\to N$ and then after considering the induced $\mathcal{O}_X$-linear map $\widetilde{M}\to\widetilde{N}$ (01I7). We note that the resulting map $\Hom_{f_*\mathcal{O}_X}(f_*\widetilde{M},f_*\widetilde{N})\to \Hom_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})$ is compatible with restrictions to distinguished open subsets $S$ and, hence, it defines a morphism of sheaves on $S$ in the opposite direction as \eqref{map-affine-equivalence-modules}. On the one hand, the composite $\Hom_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})\to \Hom_{f_*\mathcal{O}_X}(f_*\widetilde{M},f_*\widetilde{N})\to \Hom_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})$ is the identity by 01I7. On the other hand, to see that the composite $\Hom_{f_*\mathcal{O}_X}(f_*\widetilde{M},f_*\widetilde{N})\to \Hom_{\mathcal{O}_X}(\widetilde{M},\widetilde{N})\to \Hom_{f_*\mathcal{O}_X}(f_*\widetilde{M},f_*\widetilde{N})$ is the identity, it suffices to see it on the level of sheaves (i.e., with the map \eqref{map-affine-equivalence-modules} and the one in the opposite direction). In turn, it suffices to see that the composite sheaf map is the identity on sections over $D(g)\subset U$, where $g\in\Gamma(S,\mathcal{O}_S)$. Replacing $f:X\to S$ by $D(f^\sharp(g))=f^{-1}(D(g))\xrightarrow{f} D(g)$, we have that the diagram commutes.

It is left to see essential surjectivity. For this, we are going to use 01LI (note that its proof allows us to take $i_U=\text{id}_{f^{-1}(U)}$). Let $\mathcal{F}$ be a quasi-coherent $\mathcal{A}=f_*\mathcal{O}_X$-module. By previous lemma, $\mathcal{F}$ is also $\mathcal{O}_S$-quasi-coherent. For each $U\subset S$ affine, pick an $\mathcal{O}_S(U)$-module $M_U$ and fix an isomorphism $\varphi_U:\widetilde{M_U}\to\mathcal{F}|_U$ of $\mathcal{O}_U$-modules. The $\mathcal{A}|_U$-module structure on $\mathcal{F}|_U$ induces an $\mathcal{A}|_U$-module structure on $\widetilde{M_U}$ such that $\varphi_U$ promotes to an isomorphism of $\mathcal{A}|_U$-modules. Denote $M^U$ to $\Gamma(U,\widetilde{M_U})$ with the $\mathcal{A}(U)$-module structure. Then $\widetilde{M^U}$ is a quasi-coherent module over $f^{-1}(U)$. Note that $f_{U,*}\widetilde{M^U}=\widetilde{M_U}$ for each open affine $U\subset S$, where $f_U=f|_{f^{-1}(U)}:f^{-1}(U)\to U$. For each pair $U,V\subset S$ of affine open subsets with $V\subset U$, we have an isomorphism of $\mathcal{A}|_V$-modules $\varphi^U_V:\widetilde{M_U}|_V\to\widetilde{M_V}$ given by the composite $\widetilde{M_U}|_V\xrightarrow{\varphi_U|_V} (\mathcal{F}|_U)|_V= \mathcal{F}|_V\xrightarrow{\varphi_V^{-1}} \widetilde{M_V}$. The isomorphism $\varphi^U_V$ is functorial in the sense that if $W\subset V\subset U\subset S$ are affine open subsets, then $\varphi^U_W=\varphi^V_W\circ\varphi^U_V|_W$. In turn, $\varphi^U_V$ gives an isomorphism of $\mathcal{A}(V)$-modules $\widetilde{M^U}(f^{-1}(V)) =\widetilde{M_U}(V) \cong\widetilde{M_V}(V) =\widetilde{M^V}(f^{-1}(V))$. Since $f^{-1}(V)$ is affine, this determines an isomorphism of $\mathcal{O}_{f^{-1}(V)}$-modules $\theta^U_V:(\rho^U_V)^*\widetilde{M^U} =\widetilde{M^U}|_{f^{-1}(V)} \to\widetilde{M^V}$ such that $f_{V,*}\theta^U_V=\varphi^U_V$ (01I7). By the functoriality of $\varphi^U_V$ explained before and the (fully) faithfulness of $f_{W,*}$, we deduce condition (d) of 01LI. Hence, the sheaves $\widetilde{M^U}$ glue to a sheaf $\mathcal{G}$ over $X$ for which there are isomorphisms $\theta_U:\widetilde{M^U}\to \mathcal{G}|_{f^{-1}(U)}$ for each open affine $U\subset S$ such that $\theta^U_V =\theta^{-1}_V\circ\theta_U|_{f^{-1}(V)}$ for $V\subset U\subset S$ open affine. For each open affine $U\subset S$, we have an isomorphism of $\mathcal{A}|_U$-modules $\mathcal{F}|_U\cong f_*\mathcal{G}|_U$ given by the composite $$ \mathcal{F}|_U \xrightarrow{\varphi_U^{-1}} \widetilde{M}_U =f_{U,*}\widetilde{M^U} \xrightarrow{f_{U,*}\theta_U} f_{U,*}(\mathcal{G}|_{f^{-1}(U)}) =f_*\mathcal{G}|_U. $$ For $V\subset U\subset S$ open affine, we have \begin{align*} (f_{U,*}\theta_U\circ\varphi_U^{-1})|_V &=f_{U,*}\theta_U|_V\circ\varphi_U^{-1}|_V\\ &=f_{V,*}(\theta_U|_{f^{-1}(V)})\circ\varphi_U^{-1}|_V\\ &=f_{V,*}(\theta_V\circ\theta^U_V)\circ\varphi_U^{-1}|_V\\ &=f_{V,*}\theta_V\circ f_{V,*}\theta^U_V\circ\varphi_U^{-1}|_V\\ &=f_{V,*}\theta_V\circ \varphi^U_V\circ\varphi_U^{-1}|_V\\ &=f_{V,*}\theta_V\circ\varphi_V^{-1}. \end{align*} Hence, these isomorphisms glue to an isomorphism of $\mathcal{A}$-modules $\mathcal{F}\cong f_*\mathcal{G}$ (use 00AN or 009U), and we win. $\square$