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Let $X$ be an affine scheme. $U \subseteq X$ open. Then I want to show that $\mathcal{O}_X(U)$ is flat over $\mathcal{O}_X(X)$. But I want to prove it only by knowing the definition of structure sheaf as a function from $U$ to localization of $R_p$ for $p\in U$ which is locally constant. It is clear when we use the definition of structure sheaf as a localization over basic open sets. I don't want to use an equivalent definition since I couldn't extend it to the case I worked on.

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  • $\begingroup$ What is the case that you are working on? This is a special property in algebraic geometry. $\endgroup$
    – Z. M
    Commented Mar 24, 2023 at 19:57
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    $\begingroup$ I am working on an object which is called the difference scheme. Let $(R,\sigma)$ be a ring with an automorphism. The difference spectrum only contains those primes that are fixed by $\sigma$. difference scheme locally looks like this object. I need this property for proving $-\otimes _{\mathcal{O}_{X}(X)} \mathcal{O}_X(U)$ is exact. It's essential for studying quasi-coherent modules. $\endgroup$
    – Hamed Khalilian
    Commented Mar 24, 2023 at 20:23
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    $\begingroup$ I am even skeptical about this statement. I can see that it is true if you further assume that $U$ is affine. Note that, when $Z:=X\setminus U$ is of codimension 2, under some regularity, we have $\mathcal O_X(X)\cong\mathcal O_X(U)$, and this is not obvious (of course, in general, $U$ has nontrivial quasicoherent cohomology). $\endgroup$
    – Z. M
    Commented Mar 25, 2023 at 6:00
  • $\begingroup$ In general, it should be true, at least for basic open sets. Hence $R_f$ is flat $R-$module. $\endgroup$
    – Hamed Khalilian
    Commented Mar 25, 2023 at 17:10
  • $\begingroup$ An obstruction is that, although as a scheme, $U$ is the cofiltered limit of basic open subschemes of $X$ containing $U$, this limit is not compatible with taking global sections. $\endgroup$
    – Z. M
    Commented Mar 25, 2023 at 18:03

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That is not true. Let $k$ be a field, and let $X$ be $\text{Spec}(R)$ for the following $k$-algebra, $$R=k[p,q,s,t]/I, \ \ I=\langle ps,pt,qs,qt\rangle.$$ Let $U$ be the open complement of the singleton set with closed point $\langle p,q,s,t \rangle$. Then the $R$-algebra $\mathcal{O}_U(U)$ is as follows, $$\mathcal{O}(U) = R/\langle p,q \rangle \times R/\langle s,t \rangle.$$ Since the summands are not $R$-flat, also $\mathcal{O}_U(U)$ is not $R$-flat.

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  • $\begingroup$ Thanks for your answer but for example for basic open sets, it's true. Let X=spec(R), then $\mathcal{O}_X(D(f))=R_f$ which is flat module over R. Am I right? $\endgroup$
    – Hamed Khalilian
    Commented Mar 25, 2023 at 17:07
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    $\begingroup$ More generally, take $R$ to be any Noetherian S1 ring (for instance, reduced) that is not S2 (I may want excellent or at least catenary too so dimension works the way I want it to). Let $J$ denote the ideal defining the non S2-locus and let $U = Spec R \setminus V(J)$ (the complement of a codimension $\geq 2$ set). Then $R' = \Gamma(U, O_{\mathrm{Spec} R})$ is the S2-ification of $R$. $R \to R'$ is never flat if I recall correctly. For a normal or S2 ring, I wonder if it might be true (open sets that are complements of codim $\geq 2$ are fine, I worry about divisors that aren't Q-Cartier). $\endgroup$
    – Karl Schwede
    Commented Mar 28, 2023 at 0:45
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    $\begingroup$ Hamed, for your question, any localization of a ring $R$ is a flat $R$-module. $\endgroup$
    – Karl Schwede
    Commented Mar 28, 2023 at 0:48

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