For any monoid $M$, we can naturally construct a semiring $S$ as follows:


1. Let the additive monoid of $S$ be the free commutative monoid on $M$
2. Let the multiplicative monoid of $S$ be $M$


Then, if you make multiplication distribute over addition, you get a semiring.


This has an extremely simple interpretation: the underlying additive monoid can be interpreted as the set of finite multisets over the elements of $M$, with addition being just union of multisets. Then, semiring multiplication between multisets $A$ and $B$ is simply the multiset you get if you apply the original binary operation of $M$ pairwise to all elements in $A \times B$.


This construction is rather natural. Does it have a name, or is it well-known? I've found it interesting because it's arisen organically in music theory, where the semifield associated with a certain free abelian group representing musical intervals has the beautifully clear interpretation of being a semifield of musical chords.


There are a few nice variations on this idea:


1. You can instead force the free commutative monoid to be idempotent, so that it now has a natural interpretation as the set of finite subsets of $M$, rather than the set of finite multisets over it.
2. Given any semiring $S'$, you can use this construction on each of its monoids, giving you an algebraic structure with three operators that are totally ordered with respect to distributivity.


Do any of these things have names, and/or are they well-known?



Semiring naturally associated to any monoid?
It is at least sometimes called a "monoid semiring" by analogy with "[group ring](http://en.wikipedia.org/wiki/Group_ring)". As such it would be notated $S = \mathbb{N\_0}[M]$ (or $\mathbb{N}[M]$ depending how you define things).


By the way, the ring $\mathbb{Z}[M]$ you define in #1 is a commutative ring, but not a field because the element $2$ (the identity of M plus itself in $S$) has no inverse. Even if you use $\mathbb{Q}[M]$ in place of $\mathbb{Z}[M]$, you still do not necessarily get a field. For example, let $x$ denote a generator of $\mathbb{Z}$. Then $1+x\in \mathbb{Q}[\mathbb{Z}]$ has no multiplicative inverse.


