Timeline for Problem 0.9.10 in Cohn's "Free Ideal Rings and Localization in General Rings" (CUP, 2006)
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 7, 2022 at 13:43 | comment | added | Salvo Tringali | (...) $(\pi_{i,j}:H_i\to H_j)_{i,j\in\mathbb N,\,i\le j}$, where $\pi_{i,j}:=\pi_{j-1}\circ\cdots\circ\pi_i$ for $i<j$: Unless I'm missing something, $H_\omega$ is a cancellative monoid (I guess this must be a standard construction, though I don't remember to have ever seen it before); and Step 2 in the above answer comes as a special case, since starting with $H=M_\omega$ makes $H_i$ isomorphic, for each $i \in \mathbb N$, to the quotient of the free monoid on $V_\omega$ by (the smallest congruence generated by) the relations in $R_{\omega+i}$. | |
| Mar 7, 2022 at 10:55 | comment | added | Salvo Tringali | It seems to me that Step 2 is essentially a special case of the following construction: Given a monoid $H$, set $H_0:=H$ and, for each $i\in\mathbb N$, let $H_{i+1}$ be the factor monoid $H_i/\rho_i$ and $\pi_i$ the canonical map $H_i\to H_{i+1}$, where $\rho_i\subseteq H_i\times H_i$ is the smallest (monoid) congruence on $H_i$ containing all pairs $(a,b)$ s.t. $ax=bx$ or $xa=xb$ for some $x\in H_i$. Next, let $(\phi_i:H_i\to H_\omega)_{i\in \mathbb N}$ be (the canonical realization of) the direct limit (relative to the usual category of monoids) of the direct system (...) | |
| Mar 6, 2022 at 16:45 | comment | added | Pace Nielsen | Perhaps this will help. In $M$, suppose we have an equality $ab=ac$ for some words $a,b,c$ in the variables $V_{\omega}$. This equality holds by using only finitely many relations in $R_{\omega\cdot 2}$, and so all of those relations belong in $R_{\omega+n}$ for some natural number $n$. Thus, the equality $b=c$ belongs to $R_{\omega+n+1}$. | |
| Mar 6, 2022 at 8:53 | comment | added | Salvo Tringali | (Yes, sorry for the typo: I meant $u_{n+1,s,t} s = t v_{n+1,s,t}=st$.) Fine, I see your point. I haven't yet got to convince myself that $M$ is cancellative, but I'll give it another try later. | |
| Mar 6, 2022 at 3:58 | comment | added | Pace Nielsen | (Minor point: That second $u$ variable you wrote should be a $v$ variable instead.) At first, I tried to be really restrictive about which pairs of variables to allow when enlarging the relation set. But then I realized none of that mattered, for two reasons. Firstly, it was extremely difficult to control exactly when I should or should not add new variables and relations. Secondly, the cancellative property collapses all of those elements together anyway. So there was no need to make that (or any other) restriction. If $s=t$, then cancellativity makes $u_{n+1,s,t}=s=t=v_{n+1,s,t}$. | |
| Mar 5, 2022 at 21:05 | comment | added | Salvo Tringali | In the recursive def of $R_{n+1}$, do you really want to enlarge $R_n$ by adding the rels $u_{n+1,s,t} s = t u_{n+1,s,t} = st$ for all $s, t \in V_n$? Or do you rather want to add the same rels for all $s,t\in V_n$ with $s\ne t$? | |
| Mar 5, 2022 at 20:52 | comment | added | Salvo Tringali | I'll need a fresh mind to check some details, but I'm convinced that $M$ is a duo monoid: Since $V_\omega$ is a generating set of $M$, it suffices to check that, for all $s,t\in V_\omega$, there exist $p,q\in M$ s.t. $st=ps$ and $ts=sq$. To this end, fix $s,t\in V_\omega$; let $i,j\in\mathbb N$ be the smallest non-negative integers s.t. $s\in V_i$ and $t\in V_j$; and note that $st=u_{n+1,s,t}s$ and $ts=sv_{n+1,t,s}$, where $n:=\max(i,j)$. | |
| Mar 5, 2022 at 16:08 | history | answered | Pace Nielsen | CC BY-SA 4.0 |