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In the book "Algebraic Geometry I" (Gortz Wedhorn) Example 2.37 they calcul the sections of an arbitrary subset $U\subset Spec(A)=X$, ie $\mathcal{O}_X(U)$, where $A$ is an integral domain ($K=\operatorname{Frac}(A)$) :

$$\mathcal{O}_X(U)=\varprojlim_{D(f)\subset U}\mathcal{O}_X(D(f))=\varprojlim_{D(f)\subset U}A_f=\bigcap_{D(f)\subset U} A_f$$

where the intersection is take in $K=\operatorname{Frac}(A)$.

I know how to prove it directly but I don't understand how to prove it with the limit. I think it's only a commutative algebra problem...

If someone can help me I will be grateful. Also I'm sorry for my latex code (the limits is in sens of category).

Fabio Lucchini
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Kuzcohomology
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1 Answers1

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You can prove $$\varprojlim_{D(f)\subseteq U}A_f\cong\bigcap_{D(f)\subseteq U}A_f$$ by consider universal property of limits; note that you are taking a projective limit over a thin diagram of ring inclusions. Thus if $\beta_f:B\to A_f$ as $D(f)\subseteq U$, is a cone of commutative ring homomorphism over the diagram $A_f$ as $D(f)\subseteq U$, then $$A_f\subseteq A_g\implies\forall x\in B(\beta_f(x)=\beta_g(x))$$ consequently every $\beta_f$ give rise, in fact, to a ring homomorphism $$B\to\bigcap_{D(f)\subseteq U}A_f$$

Fabio Lucchini
  • 16,054
  • But in an open $U$ if there are two open $D(f)$ and $D(g)$ without $D(f)\subset D(g)$ or $D(g)\subset D(f)$, there will be no inclusion between the localized. If two open like this doesn't exist, I'm ok with your proof. – Kuzcohomology Nov 15 '19 at 13:22
  • And thank you for your answer ! – Kuzcohomology Nov 15 '19 at 13:23