2
$\begingroup$

Is there a special name or has there been any study of monoids of this form? This came up in considering the general construction of a multivariate power series algebra over a ring $R$; usually we take the set of functions $M\to R$ where $M$ is the set $[I]$ of finite multisets on the index set $I$ with pointwise addition and scalar multiplication, and define the multiplication as $(ab)_\alpha=\sum_{\beta\le\alpha} a_\beta b_{\alpha-\beta}$, where $\alpha,\beta\in[I]$, $\alpha-\beta$ is the multiset formed by removing all the elements of $\beta$ from $\alpha$ (with multiplicity), and $\beta\le\alpha$ means that all the elements in $\beta$ are in $\alpha$ with equal or greater multiplicity.

This definition makes all the monomials commute with each other, for example if $I=\{1,2\}$ then $x_1^2x_2=x_{\{1,1,2\}}=x_{\{2,1,1\}}=x_2x_1^2$ (where $x_{\{1,1,2\}}$ is notation for the coefficient function that is $1$ at $\{1,1,2\}$ and zero at every other $\alpha\in[I]$). This may not be desirable, so one can generalize to the situation where none of the variables commute, and in this case the monomials are given by strings like $x_1x_2x_1x_3$, which is to say we take $M$ to be the free monoid on $I$, with multiplication now being defined by $(ab)_\alpha=\sum_{\beta\gamma=\alpha} a_\beta b_\gamma$ (where the sum is taken over all possible values of $\beta,\gamma\in M$ such that $\beta\gamma=\alpha$). This is well defined in both the free monoid and the earlier multiset monoid (which is actually the free commutative monoid) because the equation $\beta\gamma=\alpha$ has only finitely many solutions.

So to return to the question, has there been any study or a name for monoids $M$ such that $\{\beta\in M:\beta\mid\alpha\}$ is finite for each $\alpha$, or the power series algebras that result on such monoids (by repeating the above construction on an arbitrary such monoid instead of a free monoid or free commutative monoid)?

$\endgroup$
  • $\begingroup$ Nice question. Another example is the monoid $(\mathbb{N},\times,1).$ $\endgroup$ – goblin Nov 7 '15 at 4:01
  • $\begingroup$ Related. $\endgroup$ – goblin Nov 7 '15 at 4:11
  • $\begingroup$ @goblin Wow, it's creepy how similar our questions and thought processes are here. I'd close this as a dupe if your question had an answer. $\endgroup$ – Mario Carneiro Nov 7 '15 at 6:11
  • $\begingroup$ Great minds think alike :) $\endgroup$ – goblin Nov 7 '15 at 14:34
  • 1
    $\begingroup$ There seems to be a connection here to bornological sets. If we adjoin to my "cofilter axioms" the requirement that every singleton set be small, or equivalently, that the small sets cover the monoid $M$, then we obtain the definition of a bornological set. My two "monoid axioms" seem to just be saying that the monoid operation is a "bounded map." I haven't checked the details so this might not be quite right. But anyway, this seems to suggest that the finiteness axiom is the only one with any real "teeth" or substance. $\endgroup$ – goblin Apr 18 '16 at 12:23
1
$\begingroup$

I suppose you define division in a monoid $M$ as follows: $v$ divides $u$ if there exist $x, y \in M$ such that $u = xvy$. In semigroup theory, this property is usually denoted $u \leqslant_\mathcal{J} v$ (note the inversion: $v \mid u$ is the same as $u \leqslant_\mathcal{J} v$). The relation $\leqslant_\mathcal{J}$ is one of the Green's preorders.

A monoid is said to be finite $\mathcal{J}$-above if, for each $u \in M$, the set $\{v \in M \mid u \leqslant_\mathcal{J} v\}$ is finite. I am not sure whether this was the earliest occurrence, but the term appeared in [1] and has been used several times in the literature since then.

Formal power series have also be defined on graded monoids. See [2].

[1] K. Henckell, S. Lazarus, J. Rhodes, Prime decomposition theorem for arbitrary semigroups: general holonomy decomposition and synthesis theorem. J. Pure Appl. Algebra 55 (1988), 127--172.

[2] Sakarovitch, Jacques. Chapter 4: Rational and recognisable power series. Handbook of weighted automata, 105--174, Monogr. Theoret. Comput. Sci. EATCS Ser., Springer, Berlin, 2009.

$\endgroup$
  • $\begingroup$ Actually, I define the divides relation so that $v$ divides $u$ if there exists $x$ such that $vx=u$. (It is an asymmetric operation, so you could also call it "left division".) I made a mistake in the above characterization, I thought that $vx=u$ has a unique solution $x$ if it exists, in which case if $u$ has finitely many divisors you could conclude that $vx=u$ has finitely many solutions $(v,x)$ (which is what is needed for the well-definedness of multiplication). In fact such an element $x=u/v$ is uniquely defined in both the free monoid and free comm. monoid, but not in general. $\endgroup$ – Mario Carneiro Oct 5 '15 at 17:30
  • $\begingroup$ Then your relation is the Green's preorder $\leqslant_{\mathcal{R}}$... $\endgroup$ – J.-E. Pin Oct 5 '15 at 17:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.