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Let $\mathsf{Noeth}$ be the category of noetherian rings, viewed as a full subcategory of the category $\mathsf{CRing}$ of commutative rings with one.

Let $A$ be in $\mathsf{CRing}$.

Question 1. Is there a functor from a small category to $\mathsf{Noeth}$ whose limit in $\mathsf{CRing}$ is $A$?

(I know that there is a functor from a small category to $\mathsf{Noeth}$ whose colimit is $A$.)

Let $f:A\to B$ be a morphism in $\mathsf{CRing}$ such that the map $$ \circ f:\text{Hom}_{\mathsf{CRing}}(B,C)\to\text{Hom}_{\mathsf{CRing}}(A,C) $$ sending $g$ to $g\circ f$ is bijective for all $C$ in $\mathsf{Noeth}$.

Question 2. Does this imply that $f$ is an isomorphism?

Yes to Question 1 would imply yes to Question 2.

Question 3. Does the inclusion functor $\iota:\mathsf{Noeth}\to\mathsf{CRing}$ commute with colimits? That is, if $A\in\mathsf{Noeth}$ is the colimit of a functor $\alpha$ from a small category to $\mathsf{Noeth}$, is $A$ naturally isomorphic to the colimit of $\iota\circ\alpha$?

Yes to Question 2 would imply yes to Question 3, and yes to Question 3 would imply that many colimits, and in particular many binary coproducts, do not exist in $\mathsf{Noeth}$: see this answer of Martin Brandenburg.

One may try to attack the first question as follows:

Let $A$ be in $\mathsf{CRing}$ and $I$ the set of those ideals $\mathfrak a$ of $A$ such that $A/\mathfrak a$ is noetherian. Then $I$ is an ordered set, and thus can be viewed as a category. We can form the limit of the $A/\mathfrak a$ with $\mathfrak a\in I$, and we have a natural morphism from $A$ to this limit. I'd be interested in knowing if this morphism is bijective.

Pierre-Yves Gaillard
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  • I guess by question 1 you mean if there's a functor $F$ from a small category to ${\bf Noeth}$ such that $A$ is the limit of $EF$. ($E:{\bf Noeth} \rightarrow {\bf CRing}$ is the inclusion functor.) – sqtrat Feb 01 '19 at 13:24
  • @sqtrat - Thanks! Yes. I've added "in $\mathsf{CRing}$". I hope it's clear enough now. – Pierre-Yves Gaillard Feb 01 '19 at 13:41
  • Doesn't question 2 follow from the fact that a ring is colimit of noetherian rings ?(plug in $C_i$, where $\varinjlim C_i =A$ to get $id_A$ has an antecedent, then plug in $D_i$ where $\varinjlim D_i = B$ to get that the inverse we get was a $2$-sided inverse) (I used $\varinjlim$ to denote the colimit, because in fact we can choose the colimit to be a "direct limit") – Maxime Ramzi Feb 01 '19 at 14:12
  • @Max - Sorry I don't understand. For the first step, here is how I try to follow your argument. (Let me use lim and colim.) You replace (I believe) $A$ and $C$ by $\text{colim }C_i$ with $C_i$ noetherian. You get $$\circ f:\text{Hom}(B,\text{colim }C_i)\to\text{Hom}(\text{colim }C_i,\text{colim }C_i).$$ This gives a map $$\phi:\text{Hom}(B,\text{colim }C_i)\to\lim\text{Hom}(C_i,\text{colim }C_i).$$ Let $c_i:C_i\to\text{colim }C_i$ be the coprojection. Then $$(c_i)\in\lim\text{Hom}(C_i,\text{colim }C_i)$$ corresponds to the identity of $\text{colim }C_i$... – Pierre-Yves Gaillard Feb 01 '19 at 16:05
  • You're claiming, it seems to me, that $(c_i)$ is in the image of $\phi$. Is that right? – Pierre-Yves Gaillard Feb 01 '19 at 16:05
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    @Pierre-YvesGaillard : sorry I actually wrote it down and it didn't work – Maxime Ramzi Feb 01 '19 at 16:53
  • @Max - Don't worry! It happens to me all the time! – Pierre-Yves Gaillard Feb 01 '19 at 17:21
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    Crossposted on MathOverflow: https://mathoverflow.net/q/323136/461 – Pierre-Yves Gaillard Feb 13 '19 at 12:52

1 Answers1

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Just to put this off the unanswered list:

Laurent Moret-Bailly has shown at mathoverflow that the answer is "No" for all three questions here.

Martin Brandenburg
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