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Let $R$ be a regular local complete (with respect to the maximal ideal) ring with field of fraction $K$. Let $S\cong R[[x_1,\cdots, x_n]]/J$ (this is a Noetherian local ring which is an $R$-algebra) for some ideal $J$. Assume depth$(S)>0$. Is the completed tensor product (https://stacks.math.columbia.edu/tag/0AMU) $S \widehat{\otimes}_R K$ a Noetherian local ring containing $K$ ?

Thoughts: Let $\mathfrak m$ be the unique maximal ideal of $S$. Since $K$ is a field, so the completed tensor product $S \widehat{\otimes}_R K$ should just be the inverse limit of the system $\dfrac{S\otimes_R K}{\mathfrak m^n \otimes_R K}$. I have no idea how to analyze this.

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    $\begingroup$ $S \otimes_R K$ is a localization of $S$, right? Hence $S \otimes_R K$ is Noetherian. Now I think you are simply competing a Noetherian ring along some ideal. hence its Noetherian. That said though, unless $R = K$, I think in your case you have that $m^n \otimes_R K = m^n \otimes_S (S \otimes_R K) = S \otimes_R K$. Ie, you inverted elements of $R$ that are also in the ideal $m^n$. So your completion is not very interesting unless I'm misreading things. $\endgroup$
    – Karl Schwede
    Commented Mar 3, 2023 at 15:30
  • $\begingroup$ @KarlSchwede: sorry for the late reply, but I think you are right ... I'll think about it a bit more, and then probably close my question ... $\endgroup$
    – Snake Eyes
    Commented Mar 13, 2023 at 15:48

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