Localizations of a ring at specific elements correspond to sections of a sheaf

One fundamental observation in algebraic geometry is that there are a number of powerful correspondences between ring localizations and algebraic sheaves. Although elementary, this post will outline one such correspondence, which is that the localizations of a ring to a specific element f can be exactly thought of as the sections of the sheaf of regular functions over the distinguished open set of f. (Consider this post to be a self-directed refresher! It has been a long time since I learned this material.)

Suppose that R is a finitely generated algebra over an algebraically closed field k. One might naturally wonder if there is any geometric interpretation of the localization of this ring at a single one of its elements. Say we take f \in R (where f \neq 0) and consider the local ring R_f, i.e. quotients of the form g/f^n for g \in R and nonnegative integers n. If we consider f and g as polynomials to be evaluated, then these quotients only make sense if f is always nonzero. We can in fact show that the local ring R_f corresponds exactly to the sections of the sheaf \mathcal{O} over the distinguished open set D(f).

Consider the natural homomorphism \phi : R_f \to \mathcal{O}(D(f)) which maps a formal fraction g/f^n \in R_f to the regular function which is given by the actual quotient of polynomials g/f^n \in \mathcal{O}(D(f)). This mapping is straightforwardly injective: suppose that g/f^n = 0 on D(f); then g = 0 on D(f) and f = 0 on the complement of D(f) by assumption, so fg = 0 everywhere. Hence the preimage of such a function g/f^n \in \mathcal{O}(D(f)) is exactly 0.

Surjectivity is less trivial. Obviously, every function of the form g/f^n \in \mathcal{O}(D(f)) is regular, but is every function in \mathcal{O}(D(f)) globally representable in the form g/f^n? Yes. Suppose that \psi : D(f) \to k is a regular function in \mathcal{O}(D(f)). I will sketch out the argument in broad strokes.

First, by the definition of a regular function, \psi is representable as a quotient of polynomials, g_a/f_a, in the neighborhood of every point a \in D(f). Distinguished open sets are in a sense the “smallest open sets” of the Zariski topology (i.e. they form a basis of the topology), so if we shrink these neighborhoods enough, they become distinguished open sets in their own right, say D(f_a) for every such point a \in D(f). These open neighborhoods together cover D(f) and, conversely, we may say that the variety V(f) is the intersection of every V(f_a) or, equivalently, the single variety V({f_a}).

We have essentially managed to glue together all our little pieces of information about the regular function \psi into a single piece of information about f, but how can we go further? Translating into the algebraic setting, we know of course that f \in I(V(f)), but here we can use the Nullstellensatz to yield a more tractable result; in particular, it tells us that f \in \sqrt{\langle f_a\rangle}, or that f (as an element of R) is a member of the radical of the ideal which is generated by finitely many elements f_a.

Now we know that f_n = \sum_a k_a f_a for finitely many a \in D(f). Let us define g := \sum_a k_a g_a, recalling from before that each g_a comes from the local representation of f as g_a/f_a near a \in D(f). Now taking any arbitrary point b \in D(f) with a local representation g_b/f_b of f, we can check algebraically that gf_b = g_bf^n and hence g_b/f_b = g/f^n. Therefore g/f^n is a valid global representation of \psi over D(f).

February 10th, 2023 | Posted in Math

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