Let $A$ be a ring. Is the sequence \begin{align} \textstyle A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}} \rightrightarrows \prod\_{\mathfrak{p}\_{1},\mathfrak{p}\_{2}} A\_{\mathfrak{p}\_{1}} \otimes\_{A} A\_{\mathfrak{p}\_{2}} \end{align} exact? Here the products are over all prime ideals of $A$.


*Thoughts:*


1. Since $A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}}$ is faithfully flat, the sequence \begin{align} \textstyle A \to \prod\_{\mathfrak{p}} A\_{\mathfrak{p}} \rightrightarrows (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \otimes\_{A} (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \end{align} is exact, so it is enough to show that \begin{align} \textstyle (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \otimes\_{A} (\prod\_{\mathfrak{p}} A\_{\mathfrak{p}}) \to \prod\_{\mathfrak{p}\_{1},\mathfrak{p}\_{2}} A\_{\mathfrak{p}\_{1}} \otimes\_{A} A\_{\mathfrak{p}\_{2}} \end{align} is injective.
2. If $A$ is an integral domain, the question is equivalent to asking that $A = \bigcap\_{\mathfrak{p}} A\_{\mathfrak{p}}$ inside its fraction field. See e.g. [this](https://math.stackexchange.com/questions/630752/an-integral-domain-a-is-exactly-the-intersection-of-the-localisations-of-a-a) and a related question [here](https://mathoverflow.net/questions/293224/localization-and-intersection).
3. Here I'm asking more-or-less whether sections of $\mathcal{O}\_{X}$ satisfies descent with respect to the morphism $\coprod\_{x \in X} \operatorname{Spec} \mathcal{O}\_{X,x} \to X$ (which is not an fpqc cover). It's at least true that morphisms of quasi-coherent sheaves do not satisfy descent for such maps, see [this](https://mathoverflow.net/questions/127362/counter-example-to-faithfully-flat-descent).



Descent for the "localizations at all primes" ring map
Let $S$ be a compact, totally disconnected topological space whose topology is not discrete.


Let $k$ be a field and let $A$ be the ring of locally constant $k$-valued functions on $S$.


Then I claim the kernel of $\prod\_{\mathfrak p} A\_{\mathfrak p} \to \prod\_{\mathfrak p\_1,\mathfrak p\_2} A\_{\mathfrak p\_1} \otimes A\_{\mathfrak p\_2}$ is the ring of $k$-valued functions on $S$ with no local constancy condition and thus is isomorphic to $A$.


To see this, note that every point of $S$ defines a prime ideal of $A$, and all prime ideals arise this way. For these prime ideals, $A\_{\mathfrak p}$ is just $k$, so $\prod\_p A\_{\mathfrak p}$ is the ring of $k$-valued functions on $S$ with no local-constancy condition.


Because these localizations are just fields, $A\_{\mathfrak p\_1} \otimes A\_{\mathfrak p\_2}$ vanishes unless $\mathfrak p\_1=\mathfrak p\_2$, in which case it is the field $k$ again, but the map to such a $k$ is subtracting the two pullbacks, i.e. subtracting the two identity maps $k \to k$, and thus vanishes.


Thus, the arrow is zero and so the kernel is again the ring of functions with no local-constancy condition, as desired.


Maybe with a Noetherian hypothesis it's true?


