# Commutative rings as co-limit of Noetherian rings?

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.

• I'm having trouble parsing any motivation for these questions. Why do you want these results? – Kevin Arlin Feb 12 '19 at 5:59

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 $$\mathbb{Z}$$).