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I have read that the procedure of forming field of fractions of a noncommutative ring $R$(...more generaly to adapt the general machinery of localizations to non commutative rings/modules) is much more involved and without additional conditions (eg the additional Ore condition is required) the localization of a noncommutative ring $R$ at a multiplicative subset $S \subset R$ - let write suggestively $R[S^{-1}]$ - might fail to exist, see eg here.

Treating $R$ as a category with one object and morphisms corresponding to ring elements, this problem appears in more general context of localizations of category $C$ by inverting naively some set of arrows $S$ facing the same problem that in such hypothetical loclized category $C[S^{-1}]$ the Homs might fail to form a honest set, but only a class. (eg, compare with Remarks in Localization chapter in Weibel's Homological Algebra; it seems that nearly every standard book on homological algebra I skimmed through so far not want to "make its hand dirty" on this issues on which I would like to gain deeper insight)

Now my concern is the following: I heard the slogan that this is mainly a set theoretic issue in sense that $R[S^{-1}]$ (or categorically the homs of $C[S^{-1}]$) might be even not a honest set if one would try the naive approach to mimic the commutative case, by declaring naively the underlying "set" - better class - $R[S^{-1}]$ to consist of of sequences $(r_1, s_2^{-1}, r_3, s_4^{-1}, r_5,...)$. ("zig zags")

Naive Question: Where actually the set /size theoretic issue occures if one tries this naive "adding zig zags" approach to construct the localization of $R$ (or category $C$) at $S$.

Of course in case of rings there pops up the algebraic issue that it would not be clear how to declare the ring multipliction on such new ring, but here I'm concerned with set theoretic issues only.

Could somebody explain why such object $R[S^{-1}]$, or homs of $C[S^{-1}]$ in general may happen to be not even a honest set but class?

Sources: eg "Introduction to Homological Algebra" by C. Weibel, 10.3.3 &10.3.6, or briefly indicated in ncatlab or wiki here or here.

Bowei Tang
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user267839
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    Not necessarily a duplicate, since it's about categories rather than rings, but since you mention the localization of categories, have you seen this question? – Chris Grossack Jun 25 '25 at 15:23
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    Extending any set by a single point "up to isomorphism" will naively create a proper class, since any object not already in the set is a candidate for adding. See, e.g., https://math.stackexchange.com/questions/1259858/why-is-the-collection-of-all-algebraic-extensions-of-f-not-a-set – Asaf Karagila Jun 25 '25 at 15:23
  • @Chris Grossack: Also, the discussion in the linked thread only mentioned that there arise a set theoretic problem, but it is not clarified why the naively constructed cadidate/object is actually a proper class, not a set. – user267839 Jun 25 '25 at 15:58
  • @Asaf Karagila: Sorry, I'm a bit stupid, could you maybe clarify that point? So far I understand you correctly turning it to formal terms your claim is that if say $S$ is a fixed set, then the object $X={ W \text{ set} \ \vert \ S \subset W, #(W \backslash S)=1}/\simeq$ is a not a set, but a proper class, where relation "$\simeq$" identifies two sets $W,U$ if there exist a set theoretic iso, $i: W \cong U$. If yes, wouldn't it be even of cardiality one, as if $W,U$ two sets with $S \subset W,U$ and $W-S={w}, U-S={u}$, then one – user267839 Jun 25 '25 at 15:59
  • could clearly construct an iso $W \to U$ by identifying $S$ and map $ w $ to $u$, so in $X$ $W$ and $U$ are the same. This suggests that I badly misunderstood your example, could you elaborate the construction of this object you propose to be a class but not a set? – user267839 Jun 25 '25 at 16:00
  • In formal terms, if you haven't specified your underlying set, then there is a proper class of candidates for that underlying set. Even if you require that the set extends your given ring, there are still "too many candidates". The point was that adding even one point to your set would already result in such proper class of options. – Asaf Karagila Jun 25 '25 at 16:01
  • @Asaf Karagila: Ah ok, then I misinterpreted what your meant by "up to isomorphism". So the object you suggested to consider in 1st comment was not my $X$ but just $Y:={ W \text{ set} \ \vert \ S \subset W, #(W \backslash S)=1}$, so without "quoting out"\identifying the isomorphic sets, right? Then I agree that such $Y$ is not a set, as inside $Y$ we consider any two properly distinct sets $W,U $ with $S \subset W,U$ as different objects even if there exist some $W \cong U$. – user267839 Jun 25 '25 at 16:42
  • (cont) But how does this fact explain the set theoretic problem with naive attempt to construct the localization ring $R[S^{-1}]$ at $S$ as to consist of all formal sequences of type $(r_1, s_2^{-1}, r_3, s_4^{-1}, r_5,...)$ with $r_j \in R, s_i \in S$. Isn't it harmless? Isn't it from set theoretical point the same as to do following: Say we have two sets $A,B$, and we want to construct a new set $C$ consisting of all sequences $(a_1, b_2, a_3, b_4, a_5,...)$ with $a_{2n+1} \in A, b_{2m} \in B$. I not see any set theoretic problems with such construction. – user267839 Jun 25 '25 at 16:46
  • (cont) Essentially it's just as partitioning the decimal sequence of reals in even and odd part, isn't it. Or do I missing the issue where such potential set theoretic problem of same nature as your example in construction of $R[S^{-1}]$ consisting of such sequences actually occure? – user267839 Jun 25 '25 at 16:52
  • "I have read", "I heard": please cite your sources. – Naïm Camille Favier Jun 25 '25 at 17:01
  • @Naïm Camille Favier: Added, thanks – user267839 Jun 25 '25 at 17:12
  • @user267839 I don't see anything in the wiki page about the Ore condition preventing set-theoretic issues (and indeed I don't see any set-theoretic issues here at all). Can you cite a specific part in that page, or in another source? – Noah Schweber Jun 25 '25 at 17:19
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    Size issues come up when localising a general category, because of the size mismatch between objects and morphisms. Rings are, in this context, viewed as one-object categories, so there is no issue. The references you provide don't seem to mention any size issues with localisations of rings. – Naïm Camille Favier Jun 25 '25 at 17:28
  • @Noah Schweber: See Weibel's book 10.3.3 & 10.3.6 for most "careful" elaboration I found so far; especially proposing Ore condition as sufficient condition to circumvent such potential size problems appearing in forming localization naively by adding all "zig zags" together. – user267839 Jun 25 '25 at 17:31
  • @Naïm Camille Favier: Could you elaborate what you mean by size mismatch between objects and morphisms? Doesn't the potential size problem occure only for morphisms, namely if we try to localize a cat $C$ at some set of maps $S$ then the naive $C[S^{-1}]$ would have still the same objects at $C$ but $Hom_{C[[S^{-1}]}(c,d)$ (for $c,d\in C$ admits "all possible" zig zags? The problem is why this could fail to be a set? – user267839 Jun 25 '25 at 17:38
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    The problem is that your "all possible zig-zags" quantify over the class of objects of $C$ (after all, a zig-zag consists of a sequence of objects of $C$ and a sequence of morphisms between them), so there is not a priori a set of them. – Naïm Camille Favier Jun 25 '25 at 17:55
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    @user267839 To be clear, I see how size issues arise in a general setting, but I don't see size issues when we're specifically looking at a ring as such. (That is, I agree with what Naim is saying here; there are no size issues for rings that I can see.) – Noah Schweber Jun 25 '25 at 18:00
  • @Naïm Camille Favier: Alrights, thank you. So such set theoretic problems could only appear when class of objects is not a set (...this excludes the case with ring as one-obj cat; there the job of Ore condition is exclusively to assure that he multiplication in localized ring behaves "nicely" and not to avoid any kind of set theoretical issues), by more or less clear observation that a class of objects parametrized by a class is itself a class, that's it, right? – user267839 Jun 25 '25 at 18:16
  • @Noah Schweber: Yes, this was indeed part of my initial confusion.(Ore condition could reapair both problems; so i wrongly conjectured that the problem was the same for ring as category and localization of categories in general. Think I understand it now; compare with me previous comment to Naïm Camille Favier; so its really the problem that the arrows are represented as sequences indexed by objects of $C$, which a priori form a proper class, so these "sequences" also form a class – user267839 Jun 25 '25 at 18:21

2 Answers2

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The localization of a noncommutative ring at any set of units certainly exists. There are no size issues; the Ore condition merely gives a nicer form for elements in the new "ring of fractions." Amusingly, while the Wikipedia page you link to is wrong, the nearly-identical paragraph on the noncommutative case at the linked page on localization of rings correctly admits that localizations always exist.

If you localize a locally small category at a class of morphisms, you might get zigzags between two fixed objects that have to pass through a large class of objects. For instance, take the large poset $P$ given by a large class $x_i$ of minimal elements and a maximum element $y.$ If we localize at all the maps $a_i:x_i\to y,$ then the result contains the class $c_{ij}=a_j\circ a_i^{-1}$ of automorphisms of $y$. By considering the universal property of the localization and the functors $p_i:P\to B\mathbb Z$, where $B\mathbb Z$ has a single object with endomorphisms $\mathbb Z,$ sending $a_i$ to $1$ and all other $a_j$ to $0$, we see all the $c_{ij}$ are distinct in the localization, which is thus not locally small. It might be informative to check that in this case, the analogue of the Ore condition does not hold.

This phenomenon clearly can't be repeated for rings, seen as one-object categories. And indeed, if you view a (small) ring as a one-object Ab-enriched category, you have available the result that 2-categories of categories enriched over a cocomplete monoidal closed category are always 2-categorically cocomplete, including closed under "coinverters", which are an abstraction of localizations.

Kevin Carlson
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There is no set-theoretic difficulty in constructing the localisation of a small category. You can even do it using the general adjoint functor theorem if you want. The point is that there is an easy upper bound on the number of zigzags in terms of the set (!) of all morphisms in the small category. The same goes for rings, which (unlike categories) are small by default.

Zhen Lin
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  • So set theoretic issues with localizations of cats can only potentially happen if class of objects is a proper class, right? (Then, in localized cat the class of homs should have roughly cardinality of sequences indexed by objects of $C$, so obviously also a proper class – user267839 Jun 25 '25 at 18:26
  • It can also happen if you have a proper class of morphisms. – Zhen Lin Jun 25 '25 at 18:27
  • ...or that way, right. But I intended to start with a category $C$ having Hom's to be sets – user267839 Jun 25 '25 at 18:29