The 'union of factors' comultiplication in a monoid ring?

Question

let $\mathbb{Z}[M]$ be 'the' monoid ring of $M$ over $\mathbb{Z}$; that is to say (if my understanding is right) the set of finite linear combinations of elements of $M$, with product given by $\sum_n i_n\mathbf{a}_n\times\sum_m j_m\mathbf{b}_m$ $= \sum_{m, n}i_nj_m(\mathbf{a}_n\times\mathbf{b}_m)$. I'm somewhat familiar with the 'standard' comultiplication that takes an element $r$ of $\mathbb{Z}[M]$ to $r\otimes r$, but I've also seen a different map $\mathbf{m}$ from $\mathbb{Z}[M]\mapsto\mathbb{Z}[M]\otimes\mathbb{Z}[M]$ that could be described as 'all possible product representations': $\mathbf{m}(r) = \sum_{\mathbf{s}, \mathbf{t}: \mathbf{s}\times\mathbf{t}=\mathbf{r}}\mathbf{s}\otimes\mathbf{t}$; in particular I feel like I've seen it on the monoid ring of the monoid of strings over some alphabet $A$. Is there a standard name for this construction, and has it been substantially studied? (And for that matter, do I even have the definition right or should I be looking at linearly extending the map from $M\mapsto\mathbb{Z}[M]\otimes\mathbb{Z}[M]$ that takes $\mathbf{a}\in M$ to $\sum_{\mathbf{b},\mathbf{c}\in M: \mathbf{b}\times\mathbf{c}=\mathbf{a}}\mathbf{b}\otimes\mathbf{c}$?) I'm trying to get a better grasp on 'abstract nonsense' particularly from a combinatorial perspective but my skills still feel weak.

Answer 1

One has to be careful here. The "standard" comultiplication does not send all elements $r \in \mathbb{Z}[M]$ to $r \otimes r$; only elements of $M$ comultiply that way. For example, if $m,m' \in M$ we have $$\Delta(m + m') = \Delta(m) + \Delta(m') = m \otimes m + m' \otimes m' \neq (m + m') \otimes (m + m').$$ Similarly, the second comultiplication formula only holds for elements of $M$ and is then extended linearly to $\mathbb{Z}[M]$.

I don't know if the comultiplication map defined by $\Delta(m) = \sum_{st = m} s \otimes t$ has a "name", but it is an example of a general construction. Namely, if $A$ is a finite dimensional associative algebra with multiplication $m : A \otimes A \to A$, then the dual $m^{\ast} : A^{\ast} \to (A \otimes A)^{\ast} = A^{\ast} \otimes A^{\ast}$ defines a coassociative comultiplication on $A^{\ast}$. If you apply this to $A = \mathbb{Z}[M]$ and identify $A^{\ast}$ with $A$ using the standard dual basis, you get the comultiplication described above.