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Start with a commutative unital ring $X$, and consider the collection $\mathcal{L}$ of all rings $R$ such that there exists a multiplicative subset $S \subset R$ where $R\left[S^{-1}\right] \cong X$. Is there necessarily an $R \in \mathcal{L}$ such that the only non-units of $R$ are $\pm 1$? I.e. is there a "best" un-localization of $X$? And is this $R$ going to be unique in any sense? Is every other ring $R' \in \mathcal{L}$ going to be some localization of $R$ too? Generally what is the structure of $\mathcal{L}$ for a given ring $X$? Is this question more interesting if we reformulate it for not-necessarily-unital rings instead?

Note that I'm aware this might be a poor question; I'm very much spit-balling ideas I've never considered before, this sort of "inverse problem" for localization, to see if it's meaningful and if folks have thought about it before.

Mike Pierce
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    I don't know the answers to your questions, but $X$ is always the localization of $X\times R$, so I don't think you can say there's a best localization. Specifically if $X,R$ are boolean rings, then both only have $1$ as a unit (given that in boolean rings $x^2=x$), so this property cam't be seen as "being the best localization" (at least not in a naive sense) – Maxime Ramzi Oct 17 '19 at 16:08
  • @Max for which multiplicative subset of $X \times R$ is $X$ a localization of $X \times R$? – Mike Pierce Oct 17 '19 at 16:27
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    The one generated by $(1,0)$ – Maxime Ramzi Oct 17 '19 at 16:32
  • @Max I wasn't even considering integral domains. Maybe this question is more nicely formulated for integral domains? And yeah, I didn't consider a ring where $1 = -1$, so maybe the definition of "best" un-localization should just require "as few units as possible". And maybe that aligns with the universal property idea in the answer below. – Mike Pierce Oct 17 '19 at 16:53
  • Well in the situation I described, there are "as few units as possible" (just the one). Indeed the universal property seems to align with what you were thinking of, but as explained there, there will not always (dare I say, not often) be such a universal unlocalization. In fact, I think the examples given there are wrong, say for $\mathbb F_p$, take $R' = \mathbb F_p^2, S'= (1,0)$ as above, then any ring morphism $\mathbb F_p\to R'$ sends $1 (\in S)$ to $(1,1) \notin S'$, so there is no morphism $(R,S)\to (R',S')$. This seems to always happen – Maxime Ramzi Oct 17 '19 at 18:18
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    You might consider the word "globalize" rather than "un-localize". This term is used in number theory and algebraic geometry, for instance in Ravi Vakil's or Eisenbud-Harris's books. (Or this question – Viktor Vaughn Oct 18 '19 at 01:03

2 Answers2

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I will only talk about unital commutative rings (and I mostly think about integral domains).

From your description, I think what you have in mind of a "best un-localization" is a universal property:

Definition: An un-localization ("unloc" for short) of a ring $X$ is a triple $(R, S, \iota)$, where:

  • $R$ is a ring;
  • $S \subseteq R$ is a multiplicative subset;
  • $\iota$ is a homomorphism from $R$ to $X$, such that it induces isomorphism between $S^{-1}R$ and $X$.

A best unloc of $X$ is an unloc $(R, S, \iota)$ which satisfies the following universal property:

For any unloc $(R', S', \iota')$ of $X$, there exists a unique homomorphism $\tau$ from $R$ to $R'$, such that $\tau(S) \subseteq S'$ and $\iota' \circ \tau = \iota$.

The best unloc, if exists, is then unique up to unique isomorphism.

Example:

  1. $\mathbb{Z}$ is the best unloc of $\mathbb{Q}$.

  2. $\mathbb{F}_p$ is the best unloc of itself.

However a best unloc may not exist in general.

For example, consider $X = k(t)$, the field of rational fields over a field $k$.

The inclusion $\tau:A = k[t] \hookrightarrow X$ identifies $X$ with the fraction field of $A$, hence gives an unloc of $X$.

Similarly, $\tau':A' = k[t^{-1}] \hookrightarrow X$ also gives an unloc of $X$.

Now if $\iota: R\rightarrow X$ is a best unloc, then the image $\iota(R)$ should be contained in both $\tau(A)$ and $\tau'(A')$, whose intersection is $k$. This of course is not possible.

WhatsUp
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  • With that last example $X = k(t)$ in mind, there's a symmetry between $k[t]$ and $k[t^{-1}]$ that makes me want to claim that the best unloc will be unique up to an automorphism of $X$, or something similar. – Mike Pierce Oct 17 '19 at 17:03
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    I guess even this is not true. The example $X = k(t)$ is the function field of the curve $\mathbb{P}^1$. In general, if you take two complete varieties which are birationally equivalent but not isomorphic, then you might get counterexamples by taking $X$ to be the function field. There is of course no example in dimension one, since birationally equivalent smooth projective curves are isomorphic. – WhatsUp Oct 17 '19 at 17:07
  • I think your examples are mistaken. Indeed take $(R,S,\iota)$ a best unloc. of $X$, and consider $R' = X\times \mathbb Z, S' = {(1,0)}, \iota'=$ the projection onto $X$. Then $(R',S',\iota')$ is an unloc. of $X$ but there is no morphism $R\to R'$ with $f(S)\subset S'$ : any $f$ has $f(1) = (1,1) \notin S'$. I can think of at least two ways of fixing this : requiring that $\iota$ be injective (this feels a bit unnatural) or restricting to $X,R,R'$ integral in all definitions – Maxime Ramzi Oct 17 '19 at 20:35
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I'm not sure about your more general question about the structure of $\mathcal L$ but any finite field not equal to $\Bbb F_2$ or $\Bbb F_3$ (or more generally any integral domain containing a finite field $F\neq\Bbb F_2,\Bbb F_3$) will give an example of a ring which does not have an "un-localization"

Alex Mathers
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