Understanding the sheaf $\mathcal{O}_x(U)$ over an open $U\subseteq X$ as a subset of the Cartesian product $\prod_{P\in U} R_P$.

I am reading from the red book of varieties and schemes of David Mumford about sheaves: Let $$R$$ be a commutative ring, $$1\in R$$, $$X=\text{Spec}R$$ the set of all prime ideals $$P \nsubseteq R$$ and a $$U\subseteq X$$ an open subset of $$X$$ with the Zariski topology. Moreover, let $$R_S=S^{-1}R$$ the localization of $$R$$ over a multiplicatively closed subset $$S\subseteq R$$.

Definition: The author defines $$\mathcal{O}_x(U)=\Gamma(U,\mathcal{O}_X)$$ to be the set of elements $$\{s_P\} \in \prod_{P\in U} R_P$$ for which there exists a covering of U by distinguished open sets $$X_{f_a}$$ together with elements $$s_a\in R_{f_a}$$ such that $$s_P$$ equals the image of $$s_a$$ in $$R_P$$ whenever $$P \in X_{f_a}$$.

Questions:

1) What exactly are the elements $$\{s_p\}$$? Is an element $$s\in \prod_{P\in U} R_P$$ a "P-tuple"? Is it something like $$s=(...,s_P,s_Q,...) \in \prod_{P\in U} R_P$$ such that $$s_P \in R_P$$, $$s_Q \in R_Q$$ for some prime $$P,Q \in U$$?

2) I suppose that to find the image of $$s_a$$ in $$R_P$$, we use the fact that $$P \in X_{f_a}$$, and so $$f_a \notin P$$, so we can write $$s_a=\frac{g_a}{{f_a}^n} \in R_{f_a}$$, hence the image of $$s_a$$ in $$R_p$$ is the corresponding germ $$[s_a]=[(\frac{g_a}{{f_a}^n},X_{f_a})]$$ in the direct limit $$R_P=\varinjlim\limits_{X_f \in U_p} R_f,$$ where $$U_p=\{X_f:P\in X_f\}$$. Is that right?

3) I can't see how $$s_P$$ can coincide with $$[s_a]$$, where $$[s_a]$$ is the image of $$s_a$$ in $$R_P$$. Can you give me a simple example of a commutative ring $$R$$ and an open $$U\nsubseteq X$$ such that $$\{s_P\}_{P \in U}=\{[s_a]\}_{a\in A}$$ as in the above definition? What is $$\mathcal{O}_x(U)$$ in this case and how is it related with the tuples $$\{s_P\} \in \prod_{P\in U} R_P$$?

4) Does there exist an alternative definition of $$\mathcal{O}_x(U)$$ or a book/link which gives more details/examples about this definition?

• You've repeated the statement "$\mathcal{O}_x(U) = \Gamma(U,\Omega_X)$" several times - this should be $\mathcal{O}_X(U)=\Gamma(U,\mathcal{O}_X)$. Aug 19, 2019 at 19:45
• @KReiser Maybe you are right. However, in the book that I am reading, the notation $\Gamma(U,\Omega_X)$ is used when talking about the sheaf which I defined above. I found the notation $\mathcal{O}_X(U)$ somewhere else. Maybe is it more correct if I edit $\mathcal{O}_X(U)$ to $\Omega_X(U)$? Aug 19, 2019 at 19:55
• @bing-nagata-smirnov $\mathcal{O}_X$ is the usual name for the sheaf you are describing, and $\Omega_X$ is the usual name for something entirely different (namely, the sheaf of differential forms). I've never read Mumford; I guess he's using nonstandard notation?
– user14972
Aug 19, 2019 at 19:55
• @Hurkyl Oh, I see. Then I'll change the notation as you recommend me to do in order not to create confusions. Thank you both for the comments! Aug 19, 2019 at 19:58

To answer (4) first, if $$X = \mathrm{Spec}(R)$$, you can actually define the structure sheaf $$\mathcal{O}_X$$ to be the unique (up to isomorphism) sheaf satisfying

$$\mathcal{O}_X(X_f) = R_f$$

and for each inclusion $$X_{fg} \subseteq X_f$$, the transition map is $$R_f \to R_{fg}$$. Of course, it takes some work to show that this constructs a well-define sheaf.

As an example, the stacks project takes this approach, although they've defined a lot of machinery prior to this point.

To answer (1), I think to fully understand the idea it helps to consider the following calculation.

If you have a product of identical terms, there is a natural bijection of sets

$$\prod_{x \in S} T \cong T^S$$

That is, the $$S$$-indexed product of copies of $$T$$ can be viewed as the set of functions from $$S$$ to $$T$$. In fact, in some set-theoretic foundations, these two sets would be literally equal!

The case of $$\prod_{P \in U} R_P$$ is a bit more awkward. You can think of its elements as being functions on the domain of primes in $$U$$, but the codomain is different for each point. Infinite products like this are a standard way of expressing such an idea.

So yes, if $$s \in \prod_{P \in U} R_P$$, it is indeed a tuple whose index set is the points of $$P$$, and I imagine the notation Mumford uses will be in line with that. But for intuition, there are times where you may be better served by the interpretation of a tuple as expressing a function on the index set.

An intuition for this is that the infinite product $$\prod_{P \in U} R_P$$ can be thought of as the set of "discontinuous" functions on $$U$$, whereas $$\mathcal{O}_X(U)$$ is the subset of "smooth" functions on $$U$$ (or 'regular' or 'algebraic' or whatever informal description you want to use).

As I mentioned before, $$\mathcal{O}_X(X_f) \cong R_f$$. The idea behind this correspondence is that if $$r \in R_f$$, then we can think of $$r$$ as a function whose value at $$P \in X_f$$ is precisely the class $$[r]_P \in R_P$$.

(I've added a subscript to emphasize that $$[r]_P$$ depends on $$P$$)

For example, if $$r \in R$$, then we can define the tuple $$s$$ whose $$P$$-th component is $$s_P = [r]_P$$. Then the fact $$s \in \mathcal{O}_X(X)$$ can be seen because $$X$$ is covered by the open set $$X_1$$, and we take the section $$s_1 = r$$. Then for every point $$P$$, $$s_P = [s_1]_P$$.

Given $$P \in X_f$$, a simpler description is that the map $$R_f \to R_P$$ is just the usual localization homomorphism. Every fraction in $$R_f$$ is already of in the form of a fraction in $$R_P$$, so it does indeed send $$g/f^n \mapsto g/f^n$$.
But the filtered colimit you write is indeed a formula for $$R_P$$: $$\mathop{\operatorname{colim}}_{X_f \ni P} R_f \cong R_P$$ and the insertion map $$R_f \to \mathop{\operatorname{colim}}_{X_f \ni P} R_f$$ does indeed send $$r \in R_f$$ to the germ $$[(r, X_f)]$$ of that colimit.
If $$P$$ is a prime ideal of $$R$$, then the set of elements not in $$P$$ is a multiplicative subset, and the usual definition of $$R_P$$ is the localization where you invert the elements that are not in $$P$$.
I'm mentioning this since your exposition suggests Mumford gave a different definition of $$R_P$$; specifically, the filtered colimit you mentioned earlier.