"Right denominator set" | How to show that the relation is transitive?

This problem comes from N. Bourbaki - Algebra I - Chapter 1.


Let EE be a monoid and SES\subset E a submonoid satisfying the following properties:

(1)(1) For every aEa\in E and sSs\in S, there exists tSt\in S and bEb\in E such that sb=atsb=at.

(2)(2) For every a,bEa,b\in E and sSs\in S such that sa=sbsa=sb, there exists tSt\in S such that at=btat=bt.

Define a relation (a,s)(b,t)(a,s)\sim (b,t) on E×SE\times S to mean that there exists s,tSs',t'\in S such that tt=sstt'=ss' and as=btas'=bt'.

Show that the above \sim is an equivalence relation on E×SE\times S.


My first attempt (to show the transitivity) :

Let (a,s)(b,t)(a,s)\sim (b,t) and (b,t)(c,u)(b,t)\sim (c,u). Then

s,tS s.t. tt=ss,as=bt\exists s',t'\in S\text{ s.t. } tt'=ss',as'=bt'

t,uS s.t. uu=tt,bt=cu\exists t'',u'\in S\text{ s.t. } uu'=tt'',bt''=cu'

By (1)(1), for (t,t)E×S(t',t'')\in E\times S, there exists (v,w)E×S(v,w)\in E\times S s.t. tw=tvt'w=t''v. Thus

asw=btw=btv=cuvas'w=bt'w=bt''v=cu'v

ssw=ttw=ttv=uuvss'w=tt'w=tt''v=uu'v

Since s,wSs',w\in S and SS is a monoid, we have swSs'w\in S. But uvu'v may not belong to SS. Besides, the second condition has not been used. So I got stuck here. And my other attempts also failed because some element may not belong to SS.


Update. My second attempt (a similar method in T.Y.Lam's book Modules and Rings - Theorem 10.6) :

Let (a,s)(b,t)(a,s)\sim (b,t) and (b,t)(c,u)(b,t)\sim (c,u). Then s,tS s.t. tt=ss,as=bt\exists s',t'\in S\text{ s.t. } tt'=ss',as'=bt' t,uS s.t. uu=tt,bt=cu\exists t'',u'\in S\text{ s.t. } uu'=tt'',bt''=cu' First we try to obtain a common ``denominator''. Note that ttS,ttStt'\in S,tt''\in S, by condition (1)(1) we have ttv=ttwStt''v=tt'w\in S for some vEv\in E and wSw\in S.

Now by condition (2)(2) we have tvt^=twt^St''v\widehat{t}=t'w\widehat{t}\in S for some t^S\widehat{t}\in S.

Thus we get c(uvt^)=(cu)vt^=(bt)vt^=b(tvt^)=b(twt^)=(bt)wt^=(as)wt^=a(swt^)c(u'v\widehat{t})=(cu')v\widehat{t}=(bt'')v\widehat{t} =b(t''v\widehat{t}) =b(t'w\widehat{t})=(bt')w\widehat{t}=(as')w\widehat{t}=a(s'w\widehat{t}) Besides, we have u(uvt^)=(uu)vt^=(tt)vt^=t(tvt^)=t(twt^)=(tt)wt^=(ss)wt^=s(swt^)u(u'v\widehat{t})=(uu')v\widehat{t}=(tt'')v\widehat{t}=t(t''v\widehat{t}) =t(t'w\widehat{t})=(tt')w\widehat{t}=(ss')w\widehat{t}=s(s'w\widehat{t}) Since s,w,t^Ss',w,\widehat{t}\in S and SS is a submonoid of EE, swt^Ss'w\widehat{t}\in S. The remaining obstacle is that I cannot decide whether uvt^Su'v\widehat{t}\in S holds or not.


Any comments or suggestions are welcome.

Note by Haosen Chen
2 weeks, 6 days ago

No vote yet
1 vote

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I think you started in the right way. The only extra leverage I can see is that, since tt' and tt'' are both in S,S, you can find solutions to your equation tw=tvt'w = t''v in two ways: one with v1Ev_1 \in E and w1S,w_1 \in S, and another with v2Sv_2 \in S and w2E.w_2 \in E.

I played around with this a little--I even managed to get to a point where I used (2) in a natural-looking way--but I couldn't get an expression involving only things in S.S.

Patrick Corn - 2 weeks, 4 days ago

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