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 $A$.


  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.

1 Answer 1


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?

  • 5
    $\begingroup$ Thanks! I think you meant "and thus is not isomorphic" :) $\endgroup$ Aug 23, 2021 at 20:39

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