Problem 0.9.10 in Cohn's "Free Ideal Rings and Localization in General Rings" (CUP, 2006)

Question

Let $S$ be a monoid. On p. xvii of P.M. Cohn's Free Ideal Rings and Localization in General Rings (CUP, 2006), one reads that

• an element $u \in S$ is regular if (quote) "[...] it can be cancelled, i.e. $ua = ub$ or $au = bu$ implies $a = b$";
• $S$ is a conical monoid if (quote) "$ab = 1$ implies $a = 1$ (and so also $b = 1$)".

On p. 53, one then reads that $S$ is an invariant monoid if each of its elements is invariant, an element $c \in S$ being invariant if $c$ is regular and $cS = Sc$ (so in particular, an invariant monoid is cancellative).

Now, with these definitions in mind, Problem 0.9.10 in the same book (p. 58) asks:

Is every invariant conical monoid necessarily commutative?

The problem is numbered "10°" by Cohn, and on p. xiv one reads:

[...] open-ended (or open) problems are marked °, though sometimes this may refer only to the last part; the meaning will usually be clear.

At first, I had mistakenly thought (see the first version of this post) that the problem had a trivial answer, due to the following (flawed) argument:

Start with a semigroup $H$, pick an element $e \notin H$, and let $H^{(e)}$ be the (unique) magma obtained by extending the operation of $H$ to a binary operation on $H^{(e)}$ in such a way that $ex = xe := x$ for every $x \in H^{(e)}$. Clearly $H^{(e)}$ is a monoid (some people would call it an unconditional unitization of $H$); and it is cancellative, commutative, or invariant if and only if so is $H$, resp. On the other hand, $H^{(e)}$ is obviously conical (regardless of whether this is the case with $H$). Therefore, if $H$ is a non-commutative invariant monoid (such as the monoid of non-zero elements of a non-commutative skew-field), then $H^{(e)}$ provides a negative answer to Cohn's question.

The issue here is that $H^{(e)}$ is a cancellative monoid if and only if $H$ is a cancellative semigroup without unity: If $H$ is a monoid with identity $1_H$, then $xe = x1_H = x = 1_H x = ex$ for every $x \in H$ (and, by construction, $e \ne 1_H$). So, my (new) question is:

Q. Is anyone aware of any progress on Cohn's problem since 2006?

Update (12/03/2022). Cohn's problem has been quickly solved by Pace Nielsen below. A second and, in some sense, much easier solution was communicated by George Bergman to Pace Nielsen and is now the bulk of the accepted answer of this thread.

Answer 1

I mentioned this problem to George Bergman and he offered the following much simpler solution, which he has given permission for me to post. In his own words:

Take any nonabelian group $G$ with a nontrivial homomorphism $f\colon G \to \mathbb{Z},$ and let $S$ be the submonoid of $G$ with underlying set $\{e\} \cup \{x\in G\, |\, f(x) > 0 \}.$

If $G$ is the free group on the two symbols $x,y$, and $f$ is the homomorphism determined by the rule $f(x)=f(y)=1$, then I believe that the monoid $M$ described in my other answer is the "duo-hull" of $x$ and $y$ inside the monoid $S$.

Answer 2

Here is a solution to the problem. (I'm unsure if this problem has been solved elsewhere.)

Step 1: First construct a duo monoid in an almost free manner, starting from two generators.

To do this, let $V_0=\{x,y\}$ be a set containing two variables, and let $R_0=\emptyset$ be the collection of relations that we assume they must satisfy.

Now, assuming that $V_n$ and $R_n$ have been defined for some integer $n\geq 0$, we recursively define $$ V_{n+1} = V_n\cup \{u_{n+1,s,t},v_{n+1,s,t}\, :\, s,t\in V_n\}. $$ So, for each pair of variables $s,t\in V_n$ we have created two new variables. Moreover, we recursively define $$ R_{n+1}=R_n\cup \{u_{n+1,s,t}s=st,\ tv_{n+1,s,t}=st \, :\, s,t\in V_n\}. $$ This guarantees that variables can push past each other (to either the left or right).

Now, take $V_{\omega}=\bigcup_{n\geq 0}V_n$ and $R_{\omega}=\bigcup_{n\geq 0}R_n$. The monoid $M_{\omega}$ generated by $V_{\omega}$ subject to the relations in $R_{\omega}$ is duo.

Step 2: We next force cancellativity.

The monoid $M_{\omega}$ is not cancellative. (Indeed, we have $u_{1,x,y}x=u_{2,x,y}x$ but $u_{1,x,y}\neq u_{2,x,y}$.) So if $ab=ac$ in $M_{\omega}$, we add $b=c$ to the set of relations, and similarly if $ba=ca$, we add $b=c$ to the set of relations.

Let $M_{\omega+1}$ be the monoid generated by $V_{\omega}$ subject to the relations in $R_{\omega+1}$. It might be the case that $M_{\omega+1}$ is guaranteed to be cancellative; however, I doubt it. In any case, we can repeat the previous process, adding new relations to $R_{\omega+1}$ if necessary. After repeating this process countably many times, we have a set of relation $R_{\omega\cdot 2}$, and the monoid $M=M_{\omega\cdot 2}$ generated by $V_{\omega}$ and subject to these relations is both cancellative and duo.

It is important to note that after forcing cancellativity, this construction is now truly a free construction. There is no freedom in the introduction of the new variables forcing the duo conditions; for if $us=st$ and $u's=st$, then $u=u'$ (and similarly on the other side)

Step 3: The conical condition comes for free!

Treating each variable as having degree/grade $1$, we can check that in each of the previous two steps, all of the relations are homogeneous in the grading. Thus, in particular $M$ is conical; the only one-sided unit is $1$, by a simple degree argument

Step 4: Check that $x$ and $y$ don't commute in $M$.

We do this by noting that the ring of real quaternions, $Q$, is a domain. Thus $Q-\{0\}$ is a duo, cancellative monoid (but clearly not conical). From freeness, we can map $M$ to $Q-\{0\}$ by sending $x$ and $y$ to nonzero, noncommuting elements. Since their images don't commute in $Q$, the relations in $R_{\omega\cdot 2}$ cannot force commutativity of $x$ and $y$.

Answer 3

Let $\mathcal G = (G, \preceq)$ be a [pre]ordered group, by which we mean that (i) $G$ is a (multiplicatively written) group; (ii) $\preceq$ is a partial [pre]order on (the underlying set of) $G$ such that, if $x \preceq y$ and $u \preceq v$, then $xy \preceq uv$; (iii) the last inequality is strict whenever $x \prec y$ or $u \prec v$.

It is readily checked that the positive cone $\mathcal G^+ := \{x \in G \colon 1_G \prec x\}$ of $\mathcal G$ is a non-unital subsemigroup of $G$ that is closed under conjugation in the sense that $a^{-1}\mathcal G^+ a \subseteq \mathcal G^+$ for every $a \in G$ (if $x \in \mathcal G^+$ and $a \in G$, then $1_G = a^{-1} 1_G a \prec a^{-1} x a$); cf. the first half of Lemma 1 in

• K. Iwasawa, On linearly ordered groups, J. Math. Soc. Japan 1 (1948), No. 1, 1-9.

It follows that $\mathcal G^+$ is a non-unital, duo subsemigroup of $G$ (a semigroup $S$ is duo if $aS = Sa$ for each $a \in S$) and hence the unitization $\{1_G\} \cup \mathcal G^+$ of $\mathcal G^+$ is a conical, invariant submonoid of $G$ in the sense of the OP (every subsemigroup of a group is obviously cancellative).

Suppose now that there exist non-commuting elements $x, y \in G$ such that $1_G \prec x \preceq y$, and note that this condition is certainly satisfied if $\mathcal G$ is a non-commutative, totally ordered group, i.e., $G$ is a non-commutative group and $\preceq$ is a total order (many examples are listed in another MO thread). It is then clear that $\{1_H\} \cup \mathcal G^+$ is a non-commutative submonoid, and so we are done.

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This fixes the faulty argument originally mentioned in the OP and recovers Bergman's solution (as reported by Pace Nielsen in the accepted answer of the thread): If there is a group homomorphism $f$ from $G$ to the additive group of the integers, then taking $\preceq$ to be the pullback of the usual order $\leq$ on $\mathbb Z$ along $f$ (so that $a \preceq b$, for some $a, b \in G$, iff $f(a) \le f(b)$) makes $(G, \preceq)$ into a totally preordered group.