Timeline for Is every commutative ring a limit of noetherian rings?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2019 at 17:28 | comment | added | R. van Dobben de Bruyn | A question closely related to your last question also appeared separately here. | |
| Feb 14, 2019 at 20:33 | answer | added | François Brunault | timeline score: 13 | |
| Feb 14, 2019 at 14:11 | history | edited | Pierre-Yves Gaillard | CC BY-SA 4.0 |
edit clearly indicated
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| Feb 14, 2019 at 13:01 | vote | accept | Pierre-Yves Gaillard | ||
| Feb 13, 2019 at 20:25 | comment | added | Denis Nardin | @Pierre-YvesGaillard Sorry, I misread the question. I thought you were talking about colimits and I am apparently bad enough that I did not get it even with some other commenter already making this mistake... | |
| Feb 13, 2019 at 20:04 | comment | added | Pierre-Yves Gaillard | @DenisNardin - Here is my (perhaps incorrect) argument for the claim "yes to Question 1 would imply yes to Question 2": Let $C\in\mathsf{CRing}$ be arbitrary, write $C=\lim_iC_i$ with $C_i$ noetherian, set $\text{Hom}:=\text{Hom}_{\mathsf{CRing}}$, and note $$\text{Hom}(B,C)\simeq\lim_i\text{Hom}(B,C_i)\simeq\lim_i\text{Hom}(A,C_i)\simeq\text{Hom}(A,C).$$ This implies that $f$ is an isomorphism. Thanks for telling me what's wrong with this, and for spelling out your argument. | |
| Feb 13, 2019 at 16:02 | comment | added | YCor | Sorry yes, "colimits" should be "limits" in my second sentence of my previous comment. | |
| Feb 13, 2019 at 15:58 | comment | added | Pierre-Yves Gaillard | @YCor - You mean that "limit" and "projective limit" are synonymous, and so are "colimit" and "inductive limit", not the other way around. Don't you? | |
| Feb 13, 2019 at 15:40 | comment | added | YCor | I indeed know many people using "limits" for inductive limits, and actually with several possible conventions. It's anyway useful to specify. Even for colimits, which I understand to be projective limits, there are several interpretations according to whether one makes assumptions on the morphisms, and whether one makes assumptions on the index category (filtered?). I'm ready to understand that for some people the choice of conventions is obvious, but it's not to me. | |
| Feb 13, 2019 at 15:17 | history | edited | Pierre-Yves Gaillard | CC BY-SA 4.0 |
edit clearly indicated
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| Feb 13, 2019 at 14:58 | answer | added | Laurent Moret-Bailly | timeline score: 36 | |
| Feb 13, 2019 at 14:43 | comment | added | Laurent Moret-Bailly | Then take $A=$ the ring of all algebraic integers. Its only noetherian quotients are finite products of residue fields of maximal ideals. So the limit is the product of all those residue fields. | |
| Feb 13, 2019 at 14:20 | comment | added | David Roberts♦ | @FrançoisBrunault $\mathbb{Z}$ is Noetherian, so the ideal $\{0\}$ is such $\mathfrak{a}$ and so $I$ has an initial object, and so the limit is just $\mathbb{Z}$, no? | |
| Feb 13, 2019 at 14:09 | comment | added | François Brunault | Regarding your last paragraph, if you take $A=\mathbb{Z}$, don't you get as a limit $\hat{\mathbb{Z}}$? In this case the morphism is not bijective. | |
| Feb 13, 2019 at 13:10 | comment | added | Pierre-Yves Gaillard | @YCor - Thanks! The question is about limits, not colimits. In question 1 I wrote "Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?". Do you think it's not a clear enough definition of what is meant by "limit"? | |
| Feb 13, 2019 at 13:03 | comment | added | YCor | Every (associative unital) commutative ring is the union of its finitely generated subrings, which are noetherian, and hence is an filtering inductive limit of noetherian rings with injective homomorphisms. This maybe answers Question 4, but you haven't defined what you mean by limit. | |
| Feb 13, 2019 at 12:51 | history | asked | Pierre-Yves Gaillard | CC BY-SA 4.0 |