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.

  • $\begingroup$ 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.) $\endgroup$ – sqtrat Feb 1 '19 at 13:24
  • $\begingroup$ @sqtrat - Thanks! Yes. I've added "in $\mathsf{CRing}$". I hope it's clear enough now. $\endgroup$ – Pierre-Yves Gaillard Feb 1 '19 at 13:41
  • $\begingroup$ 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") $\endgroup$ – Maxime Ramzi Feb 1 '19 at 14:12
  • 1
    $\begingroup$ @Pierre-YvesGaillard : sorry I actually wrote it down and it didn't work $\endgroup$ – Maxime Ramzi Feb 1 '19 at 16:53
  • 1
    $\begingroup$ Crossposted on MathOverflow: mathoverflow.net/q/323136/461 $\endgroup$ – Pierre-Yves Gaillard Feb 13 '19 at 12:52

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.