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Question 1: Does there exist a small category $\mathcal J$ such that for every commutative ring $A$, there is a functor $F :\mathcal J \to \mathcal CRing$ such that $ F$ takes every object to a Noetherian ring and the co-limit of $F$ is $A$ ?

Question 2: Does there exist a locally-small category $\mathcal J$ such that for every commutative ring $A$, there is a functor $F :\mathcal J \to \mathcal CRing$ such that $ F$ takes every object to a Noetherian ring and the co-limit of $F$ is $A$ ?

In both questions, $\mathcal CRing$ denotes the category of commutative rings with unity.

In the comments to this answer Commutative ring as a direct limit of Noetherian rings , Eric Wofsey mentiones to have a negative answer to Question 1; one purpose of this question is to record his answer for Question 1.

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  • $\begingroup$ I'm having trouble parsing any motivation for these questions. Why do you want these results? $\endgroup$ – Kevin Arlin Feb 12 '19 at 5:59
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For Question 1, note that if $A$ is a colimit of a diagram, then $A$ is generated as a ring by the subrings of $A$ which are images of the objects in the diagram. In particular, if $A$ is a colimit of a functor on $\mathcal{J}$ with values in Noetherian rings, then $A$ is generated by at most $\kappa$ Noetherian subrings of $A$, where $\kappa$ is the cardinality of the set of objects of $\mathcal{J}$. So to show the answer is no, it suffices to give examples of rings which cannot be generated by any given number of Noetherian subrings.

To prove this, let $R=\mathbb{Z}[S]$ where $S$ is an infinite set of indeterminates. I claim that every Noetherian subring of $R$ is contained in $\mathbb{Z}[S_0]$ for some finite $S_0\subset S$. It follows that $R$ cannot be generated by fewer than $|S|$ Noetherian subrings.

To prove the claim, suppose $R_0\subseteq R$ is not contained in $\mathbb{Z}[S_0]$ for any finite $S_0\subset S$. We recursively construct a sequence of elements $(r_n)$ in $R_0$ with no constant term such that the chain of ideals $$0\subset (r_0)\subset (r_0,r_1)\subset\dots$$ is strictly increasing, to conclude that $R_0$ is not Noetherian. Having chosen $r_0,\dots,r_{n-1}$, let $s\in S$ be a variable which appears in some element of $R_0$ but does not appear in $r_0,\dots,r_{n-1}$ (such an $s$ exists by our hypothesis on $R_0$). Let $r_n$ be an element of $R_0$ which contains a monomial involving $s$ of minimal total degree (say, of degree $d$). By add an integer to $r_n$, we may assume $r_n$ has no constant term. I claim that $r_n\not\in(r_0,\dots,r_{n-1})$ and so $(r_0,\dots,r_{n-1})\subset (r_0,\dots,r_n)$ strictly. Indeed, consider an arbitrary element $$x=\sum_{i=0}^{n-1}a_ir_i\in(r_0,\dots,r_{n-1})$$ for $a_i\in R_0$. Note that every monomial in $x$ which contains $s$ has degree greater than $d$, since such a monomial in an $a_i$ must have degree at least $d$ and each $r_i$ has no constant term. Thus $x\not=r_n$, as desired.


For Question 2, the answer is yes. For instance, you could take $\mathcal{J}$ to be the disjoint union of all small categories. Since every ring is a small colimit of Noetherian rings, you can write any ring as a $\mathcal{J}$-indexed colimit of Noetherian rings (just send all the other components of $\mathcal{J}$ to $\mathbb{Z}$).

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