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I explained what I meant by the ambiguous term "smooth".
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Neil Epstein
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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?

Let $(R,m)$ be an excellent Noetherian local ring. Let $S$ be a smooth 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?

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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Neil Epstein
  • 1.8k
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  • 11
  • 18

A question regarding base change of a smooth algebra via completion

Let $(R,m)$ be an excellent Noetherian local ring. Let $S$ be a smooth 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?