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 $P \in X_{f_a}$.


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?

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    $\begingroup$ You've repeated the statement "$\mathcal{O}_x(U) = \Gamma(U,\Omega_X)$" several times - this should be $\mathcal{O}_X(U)=\Gamma(U,\mathcal{O}_X)$. $\endgroup$
    – KReiser
    Aug 19, 2019 at 19:45
  • $\begingroup$ @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 $\Omega_X(U)$? $\endgroup$ Aug 19, 2019 at 19:55
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    $\begingroup$ @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? $\endgroup$
    – user14972
    Aug 19, 2019 at 19:55
  • $\begingroup$ @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! $\endgroup$ Aug 19, 2019 at 19:58

1 Answer 1


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.

To answer (3)

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 $P$, $s_P = [s_1]_P$.

To answer (2)

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.


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