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Suppose I start out with the ring $\mathbb{Z}$, and call $\mathcal{C}$ the smallest collection of (commutative, unital) rings closed under the following list of operations (which I am aware has some redundancy):

  1. Localization
  2. Taking quotients
  3. Adding a variable: $A\mapsto A[X]$
  4. Completion with respect to an ideal
  5. Tensor products
  6. More generally: finite limits and colimits

Since the tensor product of noetherian rings can be non-noetherian, I'm not even sure if everything in $\mathcal{C}$ is noetherian.

  • Is there an easy-to-state characterization of which rings are in $\mathcal{C}$?
  • Can much be said about what "nice" properties rings in $\mathcal{C}$ have? (I'm interpreting "nice" pretty loosely here).

I apologize in advance if this is well-known: even an answer of the form "this is all worked out in [X]" would be appreciated.

Edit: I would also be quite happy if anything could be said about $\mathcal{C}$ if we drop condition 6.

Edit: It looks like my naive hope that the objects of $\mathcal{C}$ might be "nice" is rather hopeless. If we drop conditions 5 and 6, then certainly everything in $\mathcal{C}$ is noetherian. Here is one way in which such rings might be called "nice":

If we define $\mathcal{C}$ using only 1-4, is the isomorphism problem for $\mathcal{C}$ decidable?

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    $\begingroup$ The pullback of the evident maps $\mathbb{Z}[x]\xrightarrow{}(\mathbb{Z}/2)[x]\xleftarrow{}\mathbb{Z}/2$ is in your class $\mathcal{C}$ and is not Noetherian. I think that $\mathcal{C}$ is quite unpleasant and I doubt one can say very much about it. $\endgroup$ Mar 19, 2013 at 13:57
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    $\begingroup$ The $p$-adic integers $\mathbf{Z}_p$ are in, and hence so is $\mathbf{Z}_p\otimes_{\mathbf{Z}}\mathbf{Z}_p$ which is a really horrible non-Noetherian ring. I'm with Neil on this one. I think that this question would be much better if the OP were to actually suggest one single conjecture about $\mathcal{C}$ rather than just asking vague questions about what that conjecture might be -- perhaps he has something in mind, and there's there's no evidence that this question has an answer in its current form. $\endgroup$
    – user30035
    Mar 19, 2013 at 18:57
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    $\begingroup$ I can imagine the background of this question. Sometimes one tries to prove something for general rings / gadgets, but only finds that those gadgets are closed under various operations. Then one wants to know if one has covered many gadgets, if not all reasonable ones in some sense. $\endgroup$ Mar 19, 2013 at 19:44

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