Assume $A$ is a finitely generated algebra over $k = \bar{k}$, that is an integral domain, and assume $a \in A \setminus 0$. How do I see that the sections of the sheaf of algebraic functions $\mathcal{O}_{\text{MaxSpec}\,A}$ over the distinguished open set $D(a)$ are precisely $A_a$?
Sections of the sheaf of algebraic functions
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general-topology
differential-geometry
commutative-algebra
sheaf-theory
vector-bundles
1 Answers
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This depends on your definitions. For many authors, this is the definition of the sheaf $\mathcal O$, since they define that sheaf on basis open sets and then use some abstract statement to conclude that this actually defines a sheaf on all open sets. You can look at Vakil's notes for example.
In Hartshorne the structre sheaf is defined otherwise and then the statement becomes a theorem. This is done in chapter II.2 for the prime spectrum. The same proof works for the maximal spectrum and the reason is because a finitely generated $k$-algebra is a Jacobson ring.