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Let $X$ be a scheme. In this answer it is stated (under the condition that $X$ is integral) that $\mathcal O_X (U) = \bigcap_{x \in U} \mathcal O_{X,x}$.

Since we're taking an intersection over all points in $U$, it seems that it might be enough to take the intersection over all closed points, since (locally) closed points correspond to maximal ideals, in which case the stalk is smaller. Is this true?

I don't understand why we need that $X$ is integral for the above equality to hold, so I am worried I am overlooking something.

Earthliŋ
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    There exist schemes without closed points. See https://www.math.utah.edu/~schwede/Papers/SchemeWithoutPoints.pdf So without some constraints, the answer to your question is "no". – Function Jun 21 '22 at 11:00
  • If $X$ is an integral and Jacobson scheme, then $U_0=U\cap X_0$ (where by zero subscript we mean subset of closed points) and $\mathcal{O}X(U)=\bigcap{x\in U_0}\mathcal{O}_{X,x}$. – Elías Guisado Villalgordo Oct 28 '23 at 09:43

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The result you are asking about is true. You can look here. As for the importance of $X$ being integral, you need to ask yourself what $\bigcap_{x\in U} \mathcal{O}_{X,x}$ means. One of the very convenient aspects of integral schemes is the presence of a function field, $K(X)=\mathcal{O}_{X,\xi}$, where $\xi$ is the generic point of $X$. In particular, there are injections $\mathcal{O}_{X,x}\to K(X)$. We identify $\mathcal{O}_{X,x}$ with its image in $K(X)$ and this is exactly the sense in which $\bigcap_{x\in U}\mathcal{O}_{X,x}$ is even defined.

Alekos Robotis
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  • Thanks! It seems like the projective limit can sometimes be used to replace the intersection. Can this help here? – Earthliŋ Jun 21 '22 at 11:35
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    @Earthliŋ Projective limit ... of what diagram? There is no obvious diagram here. – Zhen Lin Jun 21 '22 at 11:41
  • @ZhenLin If $S$ is a finite set of closed points in $X$, one can define $\mathcal O_{X, S} = \varinjlim_{U\supset S}\mathcal O_X(U)$, generalizing the definition of a stalk in a single point. Then for any inclusion $S\subset T$ of finite sets of closed points in $X$, the direct limit yields a map $\mathcal O_{X,S} \leftarrow \mathcal O_{X,T}$, so one could ask whether $\mathcal O_X(U)=\varprojlim_{S \subset U}\mathcal O_{X,S}$ where the limit taken is over all finite subsets of $U$. But I'm still having trouble wrapping my head around it and like you seem to imply, this might not make sense. – Earthliŋ Jun 21 '22 at 19:30