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Mar 1, 2021 at 20:33 comment added Neil Epstein @Anonymous Perfect. Thank you! Post it as an answer (along with your demonstration of non-flatness) and I'll accept it.
Mar 1, 2021 at 20:32 comment added Anonymous 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.)
Mar 1, 2021 at 20:08 comment added Piotr Achinger Often such a map is called a "regular morphism".
Mar 1, 2021 at 19:52 history edited Neil Epstein CC BY-SA 4.0
I explained what I meant by the ambiguous term "smooth".
Mar 1, 2021 at 19:51 comment added Neil Epstein @JasonStarr I see. I guess this was a language problem. By "smooth", I meant "flat, with geometrically regular fibers". I am not assuming finite type. I will edit my question accordingly.
Mar 1, 2021 at 19:26 comment added Jason Starr "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.
Mar 1, 2021 at 19:17 comment added Neil Epstein @PiotrAchinger How do you know that $\hat{R} \otimes_R S$ is Noetherian?
Mar 1, 2021 at 18:04 comment added Piotr Achinger Note that $T$ concides with the $m$-adic completion of the Noetherian $\widehat{R}\otimes_R S$, so the answer is positive.
Mar 1, 2021 at 16:33 history asked Neil Epstein CC BY-SA 4.0