# Frank

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 Name Frank Member for 3 years Seen 54 mins ago Website Location Age
 Feb6 awarded ● Nice Question Jan5 revised Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?added 501 characters in body; edited title; added 1 characters in body Jan2 comment Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?Right, I was not aware of this difference between the definition of the formal power series in infinitely many variables (i.e. which contains $\sum x_i$) as opposed to the completion of $k[x_1,\ldots]$ at $(x_1,\ldots)$ (which does not). Now that this is clear, I suppose it's not clear whether either of the two morphisms are flat! Jan2 comment Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?I'm a bit confused about your second statement if you wouldn't mind elaborating. I am taking the completion of $k[x_1,\ldots]$ along the ideal $(x_1,\ldots)$ which is nothing other than $k[[x_1,\ldots]]$. Jan2 revised Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?added 118 characters in body Jan2 revised Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?deleted 1 characters in body Jan2 comment Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?Indeed! Or more generally, if $R$ is a regular coherent ring and $m$ a maximal ideal, is the completion $\hat{R}$ along $m$ a flat $R$-module? Jan2 asked Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$? Nov28 comment $n$-path-connected components of a varietyThe answer to question 1 is Lemma 3.9 here arxiv.org/abs/1208.4055