Let $(R,m)$ be an excellent Noetherian local ring. Let $S$ be a smooth (i.e. $R \rightarrow S$ is flat and has geometrically regular fibers) Noetherian $R$-algebra. Let $T$ be the $m S$-adic completion of $S$. Then by the universality of the tensor product construction, there is a natural map $\hat{R} \otimes_R S \rightarrow T$. My question is: Does this map have to be flat?
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asked Mar 1, 2021 at 16:33
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2$\begingroup$ Note that $T$ concides with the $m$-adic completion of the Noetherian $\widehat{R}\otimes_R S$, so the answer is positive. $\endgroup$– Piotr AchingerCommented Mar 1, 2021 at 18:04
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$\begingroup$ @PiotrAchinger How do you know that $\hat{R} \otimes_R S$ is Noetherian? $\endgroup$– Neil EpsteinCommented Mar 1, 2021 at 19:17
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2$\begingroup$ "How do you know that $\widehat{R}\otimes_R S$ is Noetherian?" If $S$ is smooth over $R$, then $\widehat{R}\otimes_R S$ is smooth over the Noetherian ring $\widehat{R}$. In particular, it is finitely presented over $\widehat{R}$. Thus it is Noetherian by the Hilbert Basis Theorem. $\endgroup$– Jason StarrCommented Mar 1, 2021 at 19:26
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2$\begingroup$ Often such a map is called a "regular morphism". $\endgroup$– Piotr AchingerCommented Mar 1, 2021 at 20:08
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2$\begingroup$ With the current definition of "smoothness" (which I agree should be renamed regularity), the answer is easily no: take $S = \widehat{R}$, so $S=T=\widehat{R}$, but the map $\widehat{R} \otimes_R \widehat{R} \to \widehat{R}$ is usually not flat. For instance, if $R$ is the local ring of $\mathbf{A}^1$ at $0$, then this map is not flat. (For instance, one can see this using cotangent complexes, though it's probably more elementary than that.) $\endgroup$– AnonymousCommented Mar 1, 2021 at 20:32
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