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I want to prove the theorem. Let $A$ be a ring and $X = \operatorname{Spec} A$. Furthermore, let $M$ be an $A$-module and $F$ an $O_X$-module that is associated to $M$. Then, for any affine open subscheme $U \simeq \operatorname{Spec} B$ of $X$ for some ring $B$, the canonical map $M \otimes_A B \to F(U)$ is bijective and $F|_U$ is associated to $M \otimes_A B$ as $B$-module.

Proof. I know that $F(U)$ is $O_X(U)$-module, and so I can think $F(U)$ is $A$ module (which is from $F$ is a sheaf send $U$ to $A$-module) and $B$ (by $O_X(U) \simeq B$) module as well.

I know that the restriction morphism $F(X)\to F(U)$ will factor through a canonical morphism of B-modules $\sigma: M \otimes_A B \to F(U)$.

I know that this a morphism will induce a morphism of $O_U$ -modules $\widetilde \sigma: \widetilde {M \otimes _A B} \to F|_U$. Also, for any $f\in A$ such that $D(f)\subset U$, the canonical map $g: A \to B$ (which is induced from the bijection of $Hom((U,O_X|_U),(X,O_X))$ and $Hom(A,B)$).

Claim: this map $g:A\to B$ induces a isomorphism $A_f\to B_f$ via localization

Finally, I have commutative diagram: $$\begin{array} MM\otimes_A B & {\longrightarrow} & F(U) \\ \downarrow & & \downarrow \\ M\otimes_A B\otimes_B B_f \simeq M\otimes_A B_f &\stackrel{σ_f} {\longrightarrow} & M\otimes_A A_f \end{array} $$ where by above isomorphism of $A_f\to B_f$, we have $σ_f$ is isomorphism, thus we get $\widetilde \sigma$ is an isomorphism locally on $U$, then we get it is an isomorphism on $U$ as well, and thus we prove the theorem. As for claim,I think it is isomorphism, since $O_X(D(f))=A_f$,and $O_X$ restrict to $U$ and consider $D(f)$,it is equal to $B_f$,so we get it is isomorphism.

Does this seem fine?

hm2020
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lee
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2 Answers2

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Here's a quicker way. Let $i:\operatorname{Spec} B\to \operatorname{Spec} A$ be the inclusion. Then $i^*F=\widetilde{M\otimes_A B}$ (ref), which has global sections $M\otimes_A B$. On the other hand, $(i^*F)(\operatorname{Spec} B)\cong F(U)$ as $i$ is an open immersion, so $M\otimes_A B\cong F(U)$.

KReiser
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  • So is my proof fine？ – lee Jan 26 '24 at 07:45
  • @lee I didn't enjoy reading it - it's too long and too unclear for my tastes. Several important statements are not really proven well, but the underlying idea of defining a morphism of sheaves and then checking it's an isomorphism locally is okay. – KReiser Jan 26 '24 at 08:07
  • Thanks a lot, your answer really expand my horizon. – lee Jan 26 '24 at 08:30
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Answer: Here is another approach using limits (as defined in Atiyah-Macdonalds book on commutative algebra). Let $M$ be any $A$-module and let $\tilde{M}:=\mathcal{M}$. You define for any open set $U \subseteq Spec(A)$

$$\mathcal{M}(U):= lim_{D(f) \subseteq U}\mathcal{M}(D(f))$$

where $\mathcal{M}(D(f)):=M_f$. Since direct limits commute with tensor products you get the following formula:

$$\mathcal{M}(U):= lim_{D(f) \subseteq U} \mathcal{M}(D(f)):=$$

$$ lim_{D(f) \subseteq U} (M \otimes_A A_f) \cong $$

$$M \otimes_A (lim_{D(f) \subseteq U} A_f) \cong M \otimes_A \mathcal{O}(U).$$

If $U:=Spec(B)$ it follows $\Gamma(U, \mathcal{O})=B$ and you get

$$\mathcal{M}(U) \cong M \otimes_A B.$$

That the above definition gives $\mathcal{M}$ the structure of a sheaf is exercise 3.23 and 3.24 in Atiyah-Madcdonald.

hm2020
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