Timeline for Is every Noetherian Commutative Ring a quotient of a Noetherian Domain?
Current License: CC BY-SA 3.0
13 events
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Sep 10, 2014 at 21:47 | vote | accept | Siddharth Venkatesh | ||
Apr 4, 2014 at 8:26 | comment | added | Ricky | Ah, yes! Merci beaucoup Jérôme ! | |
Apr 4, 2014 at 8:22 | comment | added | Jérôme Poineau | @Ricky: Assume that $F \times G$ is a quotient of some noetherian domain $R$. Since $F$ is finite and $F$ is a quotient of $R$, the result tells you that the cardinality of $R$ is at most the continuum. But then $R$ cannot have a quotient that is as big as $G$. | |
Apr 4, 2014 at 7:58 | comment | added | Ricky | Probably a stupid question, but why this result implies that $F \times G$ is not the quotient of a noetherian domain? | |
Apr 4, 2014 at 2:10 | comment | added | Siddharth Venkatesh | I think Jérôme's example (given the correctness of the Harbarter reference for the Noetherianness, which I haven't gone through yet) answers that with a terrific example. | |
Apr 3, 2014 at 8:05 | comment | added | abx | So, finally, we don't know the answer for $\Bbb{Q}\times \Bbb{C}$ ... | |
Apr 2, 2014 at 19:49 | history | edited | Joël | CC BY-SA 3.0 |
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Apr 2, 2014 at 19:20 | history | edited | Joël | CC BY-SA 3.0 |
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Apr 2, 2014 at 18:55 | comment | added | Eric Wofsey | Very nice! I am reminded of a similar cardinality constraint I found in this answer. | |
Apr 2, 2014 at 18:46 | comment | added | abx | I think so too. | |
Apr 2, 2014 at 18:12 | comment | added | Joël | @abx I think Will is wrong. The proof of Lemma 2.1 is very short and simple, and could be given here for the sake of completeness (if user46855 agrees, I can edit his answer to add that proof). Great answer by the way. | |
Apr 2, 2014 at 15:25 | comment | added | abx | Interesting, this seems to contradict Will's answer (with his notation, take $R_1=\mathbb{F}_p$, $R=R_2=$ a very big field of characteristic $p$). Who is wrong? | |
Apr 2, 2014 at 14:33 | history | answered | user46855 | CC BY-SA 3.0 |