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Let $R$ be a normal noetherian domain. Write it as intersection of discrete valued domains $\bigcap_p R_p$. Assume I have schemes $X_p\rightarrow \operatorname{Spec}(R_p)$. Via the inclusion $B_{p,p^{\prime}}:=A_p\cap A_{p^{\prime}}\rightarrow A_p$, we can consider the structure sheaf $\mathcal{O}_{X_p}$ as $B_{p,p^{\prime}}$-algebra. Assume that for any pairs of such primes $p,p^{\prime}$ we have an isomorphism $\varphi_{p,p^{\prime}}:X_p\rightarrow X_{p^{\prime}}$ as $\operatorname{Spec}(B_{p,p^{\prime}})$-schemes, satisfying the usual cocycle condition. Under which conditions is it true that we can find a scheme $X\rightarrow \operatorname{Spec}(R)$ such that $X_p$ is the base change of $X$ via the morphism $\operatorname{Spec}(R_p)\rightarrow \operatorname{Spec}(R)$??

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I think you may have slightly misstated your question. I think instead of $\phi_{p,p'}$, it is more natural to assume that there exists a scheme $X_{p,p'} \to \text{Spec}(B_{p,p'})$ and isomorphisms between $X_p$, resp. $X_{p'}$, with the corresponding base change of $X_{p,p'}$. With your current definition, even the collection of identity morphisms, i.e., $X_p = \text{Spec}(R_p)$, does not give a datum. In place of the cocycle condition, perhaps you want to insist that for every triple $p$, $p'$, $p''$, there exists a scheme $X_{p,p',p''}$ over $\text{Spec}(B_{p,p',p''})$ and isomorphisms of the base changes with $X_{p,p'}$, resp. $X_{p,p''}$, $X_{p',p''}$ that are compatible with the specified isomorphisms of the further base changes to $X_p$, $X_{p'}$ and $X_{p''}$. To me this seems like the natural condition to impose if you are trying to get a gluing result out of coproducts $\text{Spec}(R_p\cap R_{p'})$ rather than a descent result out of products $\text{Spec}(R_p\otimes_R R_{p'})$.

Since this is an "under what conditions" question, I cannot give a complete answer. However, I can give a counterexample. Let $k$ be $\overline{\mathbb{Q}}$, or any other countable, algebraically closed field. Let $R$ be $k[t]$. By Hilbert's Nullstellensatz, the set $M$ of maximal ideals $p$ of $k[t]$ is naturally bijective to the set $k$. For every subset $S\subset M$, form a datum, $$\mathcal{X}_S = ((X_{S,p})_{p\in M}, (X_{S,p,p'})_{p,p'\in M}),$$ as follows. Define $X_{S,p} \to \text{Spec}(R_p)$ to be the open subset $\text{Spec}(R_p) \setminus (\{p\}\cap S)$, i.e., either $\text{Spec}(R_p)$ if $p\not\in S$, or $\text{Spec}(R_p)\setminus \{p\}$ if $p\in S$. For any pair $p,p'$ of distinct primes, we similarly define $X_{S,p,p'}$ to be $\text{Spec}(B_{p,p'}) \setminus (\{p,p'\}\cap S)$. Then both $X_{S,p}$ and $X_{S,p'}$ are canonically isomorphic to the base change of $X_{S,p,p'}$.

If there were a scheme $X\to \text{Spec}(R)$ satisfying your condition, it would have to be $\text{Spec}(R)\setminus S$ with its natural monomorphism to $\text{Spec}(R)$. But for most choices of $S$, this is not a scheme. Indeed, there are only countably many locally closed subschemes of $\text{Spec}(R)$, but there are uncountably many choices of $S$.

There may be a result of the form you ask if you add the hypothesis that every $X_p\to \text{Spec}(R_p)$ is proper.

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