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The question is as in the title:

Is the ring $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p = \mathbb{Z}_p \otimes_{\mathbb{Z}_{(p)}} \mathbb{Z}_p$ coherent?

As shown in the related question, the ring $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is coherent, and so one cannot reduce to this case. (On the other hand $\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}_p$ is not Noetherian, and so neither is $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_p$). The flatness lemma used in the answer to the rational case does also not appear to be applicable in this case.

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  • $\begingroup$ I am tempted to consider a more general question: Let $R$ be a DVR, and let $S$ be a flat $R$-algebra whose special and generic fibers are both coherent rings. Is $S$ a coherent ring? $\endgroup$
    – David Hansen
    Commented Jul 6, 2020 at 12:22
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    $\begingroup$ Never mind, there are counterexamples to my more general question! More precisely, taking $R=\mathbf{R}[x]_{(x)}$ and $S=$germs of $C^\infty$ functions at $0 \in \mathbf{R}$ gives a counterexample. In this case, the special and generic fibers of $S$ are fields, but $S$ is not coherent. $\endgroup$
    – David Hansen
    Commented Jul 6, 2020 at 12:38
  • $\begingroup$ Do we know the global dimension of $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_p$? Is it $2$? $\endgroup$
    – Badam Baplan
    Commented Jul 20, 2020 at 17:20
  • $\begingroup$ Dear Badam, I'm not sure, but it seems to be difficult/unknown. There is a paper of Auslander which considers the case of global dimension of algebras over a field, but this is as much as a I could find. $\endgroup$
    – Drew Heard
    Commented Jul 21, 2020 at 12:33
  • $\begingroup$ @DrewHeard right, I'm not even sure what gl. dim. of $\mathbb{Q}_p \otimes_\mathbb{Q} \mathbb{Q}_p$ is! Just thinking out loud because if we knew the gl. dim. to be small we might have coherence for free or at least an easier route to checking it. $\endgroup$
    – Badam Baplan
    Commented Jul 21, 2020 at 22:40

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