I am really confused on the proof of Proposition II.5.4 on Hartshorne's algebraic geometry book. Especially, I am not sure how does the previous Lemma II.5.3 work on this proposition. Let me state the lemma first:
Lemma 5.3. Let $X=\operatorname{Spec} A$ and $f\in A$, let $\mathscr{F}$ be a quasi-coherent sheaf on $X$. Then:
a) if $s\in \Gamma(X,\mathscr{F})$ is a section whose restriction to $D(f)$ is $0$, then for some $n>0$, $f^n s=0$.
b) given a section $t\in\mathscr{F}(D(f))$, then for some $n>0$, $f^n t$ extends to a global section of $\mathscr{F}$ over $X$.
Proposition 5.4. Let $X$ be a scheme. Then an $\mathscr{O}_X$-module $\mathscr{F}$ is quasi-coherent if and only if, for every open affine subset $U=\operatorname{Spec} A$ of $X$, there is an $A$-module $M$ such that $\mathscr{F}|_U\cong\tilde{M}$.
Proof. First reduce to the case when is $X$ affine. Let $M=\Gamma(X,\mathscr{F})$, then by the adjointness, we get a map $\alpha:\tilde{M}\to\mathscr{F}$. Since $\mathscr{F}$ is quasi-coherent, we cover $X$ by $D(g_i)$s with $\mathscr{F}|_{D(g_i))}\cong \tilde{M_i}$ for some $A_{g_i}$-module $M_i$. Now the lemma, applied to the open set $D(g_i)$, tells us exactly $\mathscr{F}(D(g_i))\cong M_{g_i}$ so $M_i=M_{g_i}$.
It follows that $\alpha$, restricted to $D(g_i)$ is an isomorphism for all $i$, hence itself is an isomorphism. QED.
My Question 1). How does the lemma imply the isomorphism?
My Question 2). How does $\alpha|_{D(g_i)}$ come from exactly $\alpha$?
I tried to answer the first question with the following argument: If we define $\theta:\Gamma(D(g_i),\mathscr{F})\to \Gamma(X,\mathscr{F})_{g_i}$ by mapping $t$ to $\varphi(s)$, where $\varphi:\Gamma(X,\mathscr{F})\to \Gamma(X,\mathscr{F})_{g_i}$ is the localization map, and $s$ is the extension of $t$ by the part b) of the lemma. Then I tried to show that this is an isomorphism:
(inj) if $\theta(t)=0$, then this means, by b) of the Lemma that, there exists $s\in\Gamma(X,\mathscr{F})$ and $n>0$ such that $s|_{D(g_i)}=g_i^nt$, and $\varphi(s)=0$. But $\varphi(s)=t=0$, this completes the injectivity.
(surj) given $\frac{m}{g_i^n}\in\Gamma(X,\mathscr{F})_{g_i}$, we have $m\in\Gamma(X,\mathscr{F})$. Then consider $\theta(m|_{D(g_i)})$ we get the fraction again (but well-definedness is in doubt here).
I do not have clue on the second question. The map $\alpha$ comes from gluing, as far as I know. Then I am not so sure how to proceed.
Interpretation of the Lemma I think the interpretation of the lemma is important. However, I am not able to understand it very well. I think it isa saying something like, a quasi-coherent sheaf on affine scheme behaves like a sheaf associated to a module.
Any help is appreciated! Thanks!
Best Answer
Writing $X=\operatorname{Spec} A$, for any $g\in A$ the map of $\mathcal{O}_X$-modules $\alpha: \widetilde{M} \to \mathcal{F}$ gives a map of $\mathcal{O}_X(D(g))=A_g$-modules $\alpha_g: \widetilde{M}(D(g))=M_g \to \mathcal{F}(D(g))$ by definition of a map of sheaves (recall that a map of sheaves $\varphi:\mathcal{F}\to\mathcal{G}$ on a space $X$ is a map $\varphi(U):\mathcal{F}(U)\to\mathcal{G}(U)$ for every open $U\subset X$ so that $res_{\mathcal{G},U,V}\circ\varphi(U)=\varphi(V)\circ res_{\mathcal{F},U,V}$ for every pair of opens $V\subset U$). To check that this map is an isomorphism, we may use the lemma:
Surjectivity: suppose $t\in\mathcal{F}(D(g))$ is a section, and let $g^nt\in\mathcal{F}(X)$ be the global section guaranteed by the lemma. Then $g^nt$ is also a global section of $\widetilde{M}$ by construction, as $M=\mathcal{F}(X)$, and the section $\frac{g^nt}{g^n}\in \widetilde{M}(D(g))$ maps to $t\in\mathcal{F}(D(g))$ by construction.
Injectivity: suppose $\frac{s}{g^n}\in M_g$ maps to $0$ in $\mathcal{F}(D(g))$. Then $s\in M=\mathcal{F}(X)$ restricts to zero in $\mathcal{F}(D(g))$, as $s|_{D(g)}=g^n|_{D(g)}\frac{s}{g^n}|_{D(g)}$ and $\frac{s}{g^n}|_{D(g)}=0$ by assumption. Then $g^ms = 0$ in $M$, hence $\frac{s}{g^n}=0$ in $M_g$.
The second part of the puzzle here is that if we have an affine scheme $\operatorname{Spec} R$, two $R$-modules $M,N$, and a map $\varphi:M\to N$, then $\widetilde{\varphi}:\widetilde{M}\to\widetilde{N}$ is an isomorphism iff $\varphi$ is. To see this from the material Hartshorne has covered so far, we can use proposition II.5.2(a) which states that $M\mapsto \widetilde{M}$ is a fully faithful functor. From this, we can use proposition II.1.1 to conclude that our original map must be an isomorphism by using the stalk-wise characterization of isomorphisms.
As for the interpretation of the lemma, yes, you're basically correct - this is a key step in showing that every quasi-coherent sheaf on an affine scheme is of the form $\widetilde{M}$, and it's showing that two notable properties we expect for a sheaf of that form actually hold for all quasi-coherent sheaves.