Timeline for If the quotient of a local ring is regular, does that imply that the original ring must be regular?
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| Jul 7, 2013 at 19:13 | comment | added | user26857 | @QiL'8 Maybe I'm missing something: in your case, if, for instance, $r=1$ and $R/(x)$ is regular, then the embedding dimension of $R/(x)$ equals $\dim R/(x)=\dim R-1$. I think now it follows that $x\notin\mathfrak m^2$. (What I want to say is that in the end the elements $x_i$ turn out to be outside of $\mathfrak m^2$, although you are not assuming this.) | |
| Jul 7, 2013 at 19:12 | vote | accept | DavidWayne | ||
| Jul 7, 2013 at 19:12 | comment | added | DavidWayne | Wow, that was a fast response to this question. I guess this is a result that is very well known, that I must have just overlooked in my studies. Thank you for the answer, and for the generalization and counterexample. | |
| Jul 7, 2013 at 8:17 | history | edited | user26857 | CC BY-SA 3.0 |
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| Jul 7, 2013 at 8:11 | history | answered | user26857 | CC BY-SA 3.0 |