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Suppose I have a monoid $(M,\, \cdot,\, e)$ equipped with a monoid homomorphism $\textrm{length} : M \rightarrow \mathbb{N}_+$ into the monoid of natural numbers under addition where $e$ is the only element of length $0$.

Consider the monoid ring $R := \mathbb{Z}[M]$ and the two-sided ideal $I := \{ p \in R \mid \sum_{x \in M} p(x) = 0 \}$, which is well-defined as elements of the monoid ring have finite support.

Note that the support of any element of $I$ is either empty or contains an element of length at least $1$. Therefore, the support of any element of $I^2$ is either empty or contains an element of length at least $2$. And it follows by induction that $\bigcap_{n > 0} I^n = \{ 0 \}\ (\star)$ as no element with finite support can contain an element of length longer than every $n \in \mathbb{N}$.

Now let $\hat{R}$ be the $I$-adic completion of $R$ defined in the usual way as the ring of Cauchy sequences quotiented by those that converge to $0$. As a result of $(\star)$, we can embed any element of $R$ into $\hat{R}$ as a constant sequence and, therefore, any element of $M$ into the multiplicative monoid of $\hat{R}$.

Furthermore, each element of the form $1 - i \in R$ with $i \in I$ has an inverse in $\hat{R}$: $$ (1 - i) \sum_{n \in \mathbb{N}} i^n = \sum_{n \in \mathbb{N}} i^n - i\sum_{n \in \mathbb{N}} i^n = 1 + i\sum_{n \in \mathbb{N}} i^n - i\sum_{n \in \mathbb{N}} i^n = 1 $$ where the sum is well-defined as its sequence of partial sums is evidently Cauchy.

However, for any $x \in M$, we can write $x = 1 - (1 - x) \in R$ with $1 - x \in I$. It seems to follow, therefore, that we have constructed an embedding of $M$ into the group of units of $\hat{R}$.

Now obviously this cannot be the case as not all such monoids can be embedded in a group but I can't find where the argument breaks down. There should be some issue even if we make the further assumption that the monoid is cancellative or indeed orderable as Mal'cev constructed such a monoid that cannot be embedded into a group.

Update:

  • It is not true that $\bigcap_{n > 0} I^n = \{ 0 \}$ when $M$ is not cancellative. For example, if $M = \langle a,\, b,\, c \mid ac = bc \rangle$ then we have that $(b - a)(1 - c) = b - a - bc + ac = b - a \in I$.
  • However, it still cannot hold if we assume $M$ is cancellative because of Mal'cev's counterexample.
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  • $\begingroup$ It may be a problem that the existence of a length function with the properties you want is rare and most monoids do not possess such a thing. $\endgroup$
    – Peter Kropholler
    Sep 19, 2022 at 11:30
  • $\begingroup$ @PeterKropholler indeed, one may compare it to the group case, where (appropriately defined) any group admitting a length function is free (see Lyndon-Schupp). But I suppose that depends on how “length function” is defined. A priori it seems any such monoid is at least cancellative. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Sep 19, 2022 at 11:35
  • $\begingroup$ Why does the support of every nonzero element of $I$ contain an element of lenght at least $1$? Do you assume that the only element of lenght zero is the identity? I think that you should be clearer about what you mean by "lenght homomorphism". $\endgroup$
    – Antoine Labelle
    Sep 19, 2022 at 11:41
  • $\begingroup$ The monoid <{a, b, c} | ab = ac> cannot be embedded in a group precise because it isn't cancellative and yet has a length function. The case for a group will be different because we'd need the length to be over the integers but then we can consider the sub-monoid of positive elements and replicate the argument. And indeed the integers are a non-free example of such a group. $\endgroup$
    – user491484
    Sep 19, 2022 at 11:59
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    $\begingroup$ @EddieJones But the problem still remains -- every monoid admits a homomorphism into the natural numbers under addition, just map everything to $0$. You are assuming more than you are stating. $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Sep 19, 2022 at 12:20

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