Timeline for Is $R=k[x_1,\ldots]\to k[[x_1,\ldots]]$ a flat morphism? What about $R\to\hat{R}$?
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| Jan 5, 2013 at 12:03 | history | edited | Frank | CC BY-SA 3.0 |
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| Jan 3, 2013 at 7:06 | comment | added | anon | @ayanta: I agree that these valuation rings are far from the actual question, and was only responding to the one in the comments. I also do not know what "coherent regular rings" are. Random poking on the internet (such as section 4 of s.web.umkc.edu/segal/papers/coh.pdf) suggests that "finitely generated $R$-submodules of finite free $R$-modules have a finite free resolution" is one definition (though certainly the second "free" should be replaced by "projective" to have the right notion even in the noetherian (global) case), whence my comment. | |
| Jan 3, 2013 at 6:42 | comment | added | user30180 | @unknown (google): I'm not sure what "coherent regular ring" means (it cannot be "coherent ring that is regular", since regularity of a commutative ring includes a noetherian condition in my experience), but valuation rings of complete rank-1 valued fields are coherent as rings, so for example the valuation ring of $\mathbf{C}_p$ is of the type you indicate. But such examples feel a bit removed from the nature of the question that is posed. | |
| Jan 3, 2013 at 4:08 | comment | added | anon | Regarding the general question about coherent regular rings: if $R$ is a valuation ring with a divisible value group (such as the ring of Puiseux series) and maximal ideal $m$, then the $m$-adic completion of $R$ is just $R/m$ since $m^2 = m$, which is certainly not $R$-flat. Does such an $R$ constitute a coherent regular ring? It has the property that finitely generated submodules of $R^n$ are free (since they are flat, hence finitely presented, and hence projective). | |
| Jan 3, 2013 at 0:59 | answer | added | user30180 | timeline score: 21 | |
| Jan 2, 2013 at 22:28 | comment | added | Frank | 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! | |
| Jan 2, 2013 at 22:11 | comment | added | Frank | 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 $k[[x_1,\ldots]]$. | |
| Jan 2, 2013 at 22:09 | history | edited | Frank | CC BY-SA 3.0 |
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| Jan 2, 2013 at 20:18 | comment | added | YCor | The question should be written inside the message as well, not only in the title (and use quantifiers. what is $k$? finitely many indeterminates?). See mathoverflow.net/questions/ask. | |
| Jan 2, 2013 at 19:27 | history | edited | Frank | CC BY-SA 3.0 |
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| Jan 2, 2013 at 19:24 | comment | added | Frank | 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? | |
| Jan 2, 2013 at 19:08 | comment | added | the L | Just a comment: the situation you are considering is very bad. Not only the ring is not noetherian, but the ideal $m$ that you complete with respect to is not finitely generated. In this case, assuming $k$ is a field, the $m$-adic completion of the left hand side will not be $m$-adically complete! | |
| Jan 2, 2013 at 19:00 | history | asked | Frank | CC BY-SA 3.0 |