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• CommentAuthorMartin B.
• CommentTimeJan 8th 2012

One of my first questions here on mathoverflow

http://mathoverflow.net/questions/9921/equality-of-elements-in-localization-via-universal-property-unsolved

was not clear enough; Tom Leinster wrote a long comment as an answer, not really answering my question. Then I cast a bounty and since nobody else replied, the answer by Tom Leinster was accepted automatically. Of course this means that other readers don't pay any attention anymore to this question.

Today the question is still unsolved. Is it appropriate to start a new thread with basically the same question and some more explanation? What do you think?
1.
That's an interesting quirk isn't it. I wonder what happens if Tom deletes his answer. Perhaps he can't do this.

Can you delete the _question_? If so, you could delete it and re-post it, and then presumably people won't be able to close as duplicate with a pointer to the
first question. And then if Tom or someone complains that you deleted his answer, you could just undelete the question. By this stage the new question will have been on the front page for a while and hence will "enter into the MO psyche" again, so you may well have got some new comments anyway :-)
2.

Yet another reason that offering a bounty is a bad idea...

3.

There is a "delete" button under my answer, so presumably I can delete it. I don't mind doing so if it will help Martin to get the answer he wants. Just let me know, Martin.

(In general, I'm against deleting nontrivial information from the site. You never know what might be helpful for someone in a curious way that you'd never predict. It's sad if they go looking for it later, only to find — or rather, not find — that it's disappeared. But perhaps in this case it's the best way forward.)

4.

Wouldn't it be simpler to just ask a new, better written question? If you're worried about running afoul of duplicate hunters, just include a note saying that the bounty on the old question complicates things, so you're making fresh start.

5.

I agree with Kevin's comment as to the best solution. Rather than trying to fix the old one, ask a new one but explain the history.

6.
I also agree with Kevin. Maybe include a link to this discussion as well.
• CommentAuthorquid
• CommentTimeJan 8th 2012

Regarding the general matter I agree that reasking with a note seems the best/easiest way to go. But perhaps one then should close (not delete) the original as 'no longer relvant' and/or also leave a comment on the original (perhaps not very likely, but say the new one gets answered and then for whatever reason in a year somebody stumbles over the original, being unaware of the more recent version, and answers the original; might be slightly unfortunate).

Regarding Kevin Buzzard's question and Tom Leinster's response: Tom Leinster could not delete the answer despite the button being visible. In general accepted answers cannot be deleted by users. In a related discussion I tested this (it is not a risk, since if it worked one could undo it by oneself). I believe, but on this detail I am not sure, if one clicks one even gets a dialog box asking 'really delete' but still eventually the software refuses to deleted on the grounds it being an accepted answer.

This is seems analogous with the situation for voting to close. Even after one already voted to close, the button close stays visible, only when one clicks it again, the software complains one already voted. [I assume the reason for this is that one wants to avoid to make many lookups in the database what a specific user already did to a post just when they look at it (again); for common things like votes, it is done, but for the rarer events it seems not; or it was an oversight by the developpers or still something else.]

7.

We could close the old one as a duplicate of the new one. That would automatically set up a link to the new one.

8.

@François - nice one!

9.

Actually, a better idea would be to create a new question and merge the old one with the new one. I believe that will effectively erase the accidental accept.

10.

It may be a good idea to test this approach on faketestsite first. I've put a bounty on this question. Could somebody please post an answer?

11.

btw, before a bounty expires, a moderator can remove it. It's not a good idea to get in the habit of putting up a bounty and then asking a mod to remove it, but I don't have a problem removing bounties occasionally if something like this is about to happen.

12.

Works beautifully! The new question gets all the answers and comment from the old one (though not the votes), and the answer gets unaccepted.

@Martin: please repost your question, and I (or another mod) will merge the old one into it.

• CommentAuthorMartin B.
• CommentTimeJan 28th 2012

Sorry for not answering so long.

What does it mean to merge two questions?
13.

The answers of the old question will be moved over to the new question and the old question will be deleted. All the answers will keep their point scores but the accept will be removed from the accepted answer.

• CommentAuthorMartin B.
• CommentTimeJan 28th 2012 edited

Here is the new question:

http://mathoverflow.net/questions/86923/elements-in-a-localization-category-theoretic-approach

I don't think that merging is a good idea. Tom's answer contains the definition of a localization, which is now also included in my question. It was always meant as an attempt in order to understand my question.

In the new version I've made quite some effort to clarify the question. But regarding the comments and (deleted) answers so far, there is still some confusion left. On the other hand I could solve these confusions just by pointing to some part of my question. It seems to me that some people don't read it carefully enough and then complain that the question does not make sense etc. I hope that the old story does not start over again.

I suggest that we discuss basic understanding problems here on META. If there is anything which needs to be clarified, it can be added to the question after this discussion.
14.

Martin, do you want to merge the questions or keep them separate?

• CommentAuthorMartin B.
• CommentTimeJan 29th 2012

I would like to keep them separate.
• CommentAuthormarkvs
• CommentTimeJan 29th 2012 edited

I still do not understand the purpose of the question and the question itself. It looks like nobody else (maybe including the OP) understands it well enough to make it clear to others. Am I right and the goal of the question is to construct a domain which is not a field, whose fraction field is the field with 1 element, avoiding using groups with 10 generators, 27 relations and undecidable word problem? I wonder how many elements that domain will have. 1/2?
• CommentAuthorMartin B.
• CommentTimeJan 29th 2012

No this is not the goal ...
15.
My reading of the question, after the embedded simplifications is: Can one infer directly from the universal characterization of $S^{-1}A$ that the only elements of $A$ that become 0 in $S^{-1}A$ are those that are annihilated (in $A$) by an element of $S$ ? The meta-question is what should be meant by "directly"; it should certainly prohibit going through the usual construction of $S^{-1}A$ and verifying that it works. I think what Martin would like is, given an $a \in A$ not annihilated by anything in $S, to exhibit some ring$B$and a homomorphism$A \to B$that inverts$S$but doesn't annihilate$a$. And it would be cheating to use$S^{-1}A$itself as$B$. What's wanted is (I think) something analogous to the use, mentioned in the question, of symmetric groups in dealing with pushouts of groups. • CommentAuthormarkvs • CommentTimeJan 29th 2012 @Andreas: This is more understandable but how a universal construction can give any info about elements? Isn't the sole purpose of all universal constructions to avoid looking at elements? In any case the standard proof of the usual properties of$S^{-1}A$is trivial. Why different proof? There must be some goal beyond the commutative ring case. • CommentAuthorquid • CommentTimeJan 29th 2012 @markvs: the proof and construction are certainly not complicated, but they are not really elegant either. As Martin mentions, there is this extra t, and then if one really writes down proofs of associativity and everything this is a bit tedious. • CommentAuthormarkvs • CommentTimeJan 29th 2012 @quid: The extra "t" is due to Grothendieck, I think (the Grothendieck group of a commutative monoid). It appears naturally when one tries to prove transitivity of the equivalence relation on fractions. Still I do not understand how universal constructions can possibly help explaining the "t" and even why the explaining is needed. We do not explain 23452*34321= 804896092 by using "universal constructions" (we use the good old multiplication table, even though the computation is tedious). Anyway, I am not sure that this discussion leads anywhere, and I simply wish the OP good luck with his question. • CommentAuthorMartin B. • CommentTimeJan 30th 2012 edited @Andreas: Yes this is what I'm asking for. An indirect proof seems to be unavoidable. I agree that the phrasing "directly" does not belong (yet) to formal mathematics. @markvs: "how a universal construction can give any info about elements?" - Please read the question carefully. I have given several examples where the universal property reveals information about the elements. "In any case the standard proof of the usual properties of$S^{-1}A\$ is trivial." Yes, we all agree with that. But this is not the question. I don't want to prove this again as usual, I want to *apply some method* in this context. This type of mathematics has not the purpose of discovering new theorems, but rather to *understand* fiddly arguments in a more conceptual way. But as I've already mentioned in the question, I don't claim at all that my approach is the best one. In fact, I don't care at all if it is the best one. The question does not ask for the best proof. It asks for a proof within the category theoretic framework which I have outlined quite in detail - though not in a formal way (although I've added a kind of formalization today, building on Moosbrugger's comments).

"Why different proof?" I have explained this in the question (section "Motivation").

"There must be some goal beyond the commutative ring case." See again the section "Motivation", in particular the last paragraph. Let me add the following:
Two years ago I've started to write a book about basic concepts of algebra. I don't wanted to use the usual approaches (otherwise there wouldn't be any motivation for a new book).
Namely I wanted to construct all these familiar universal objects (direct sums, tensor products, free gadgets, localizations, ...) by SAFT, i.e. general categorical arguments, and then derive the structure of the elements of these universal objects by means of their universal property. Of course this has the advantage that one does not have to verify anything in order to see that the resulting set of elements is - in fact - enrichable to a universal object. In fact, sometimes this is quite fiddly (it starts with, say, free groups). Surprisingly this worked out very well, but then localizations resisted to work :). I don't know yet if I will ever finish this book, but anyway, I would like to solve this problem for localizations. Ah by the way, noncommutative localization might be another possible application. But probably this will be even more complicated.

"The extra "t" is due to Grothendieck, I think (the Grothendieck group of a commutative monoid). It appears naturally when one tries to prove transitivity of the equivalence relation on fractions. " Yes, we know that and KConrad has already reminded us of that in the comments to the question. I've already made some objection to this argument in the comments, but let me elaborate it:

So let's be dumb and define (x,s) and (x',s') to be equivalent iff s'x=sx'. In order to show transitivity, let also (x',s') be equivalent to (x'',s''), so that s''x'=s'x''. Now we "have to" deduce s''s'x=s''sx'=ss''x'=ss'x'', i.e. we only have s''x=sx'' up to multiplication with s'. So this motivates us to "stabilize" the relation, define (x,s) and (x',s') to be equivalent iff ts'x=tsx' for some t in S. But then: a) We have to verify that this is in fact an equivalence relation. b) We have to verify that the addition of fractions is well-defined. c) The same for multiplication. d) Verify all the ring axioms. e) Verify the universal property. So yeah, it's great to know where this extra t comes from, but it does not help us at all to reduce the computations!

A more conceptual definition of the equivalence relation would be the one *generated* by the first one. Then of course a) can be omitted, and I think b),c),d),e) become considerably easier: We can then always argue with the "naive" criterion for equality of fractions and use these calculation we know from elementary school. But still we don't get these computations for free. Another disadvantage in this approach: Additional arguments are necessary in order to compute the kernel of \tau, i.e. to get the real criterion for equality of fractions.

But my question is not about possible constructions of the localization. The best one is the one using quotients of polynomial rings (see also the note by Felipe Voloch "rings of fractions the hard way" for a detailed account and Durov's definition of localizations of generalized rings), because here you construct them from well-known rings.

I hope this helped to clarify the question. If not, probably I would have to cook up a more abstract context where you cannot define objects by equivalence relations, thereby restricting the scope of possible constructions of objects ... hopefully I can avoid this :).