# Is every commutative ring a limit of noetherian rings?

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.

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