"Right denominator set" | How to show that the relation is transitive?
This problem comes from N. Bourbaki - Algebra I - Chapter 1.
Let E be a monoid and S⊂E a submonoid satisfying the following properties:
(1) For every a∈E and s∈S, there exists t∈S and b∈E such that sb=at.
(2) For every a,b∈E and s∈S such that sa=sb, there exists t∈S such that at=bt.
Define a relation (a,s)∼(b,t) on E×S to mean that there exists s′,t′∈S such that tt′=ss′ and as′=bt′.
Show that the above ∼ is an equivalence relation on E×S.
My first attempt (to show the transitivity) :
Let (a,s)∼(b,t) and (b,t)∼(c,u). Then
∃s′,t′∈S s.t. tt′=ss′,as′=bt′
∃t′′,u′∈S s.t. uu′=tt′′,bt′′=cu′
By (1), for (t′,t′′)∈E×S, there exists (v,w)∈E×S s.t. t′w=t′′v. Thus
as′w=bt′w=bt′′v=cu′v
ss′w=tt′w=tt′′v=uu′v
Since s′,w∈S and S is a monoid, we have s′w∈S. But u′v may not belong to S. 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 S.
Update. My second attempt (a similar method in T.Y.Lam's book Modules and Rings - Theorem 10.6) :
Let (a,s)∼(b,t) and (b,t)∼(c,u). Then
∃s′,t′∈S s.t. tt′=ss′,as′=bt′∃t′′,u′∈S s.t. uu′=tt′′,bt′′=cu′
First we try to obtain a common ``denominator''. Note that tt′∈S,tt′′∈S, by condition (1) we have
tt′′v=tt′w∈S
for some v∈E and w∈S.
Now by condition (2) we have
t′′vt=t′wt∈S
for some t∈S.
Thus we get
c(u′vt)=(cu′)vt=(bt′′)vt=b(t′′vt)=b(t′wt)=(bt′)wt=(as′)wt=a(s′wt)
Besides, we have
u(u′vt)=(uu′)vt=(tt′′)vt=t(t′′vt)=t(t′wt)=(tt′)wt=(ss′)wt=s(s′wt)
Since s′,w,t∈S and S is a submonoid of E, s′wt∈S. The remaining obstacle is that I cannot decide whether u′vt∈S holds or not.
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Comments
I think you started in the right way. The only extra leverage I can see is that, since t′ and t′′ are both in S, you can find solutions to your equation t′w=t′′v in two ways: one with v1∈E and w1∈S, and another with v2∈S and w2∈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.
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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
I think you started in the right way. The only extra leverage I can see is that, since t′ and t′′ are both in S, you can find solutions to your equation t′w=t′′v in two ways: one with v1∈E and w1∈S, and another with v2∈S and w2∈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.