# Descent for the "localizations at all primes" ring map

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 and a related question here.
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.

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?

• Thanks! I think you meant "and thus is not isomorphic" :) Aug 23, 2021 at 20:39