# What properties of $R$ does the monoid ring $R[M]$ inherit?

It's known that polynomial rings $$R[x_1,...,x_n]$$ inherit some properties of a base ring $$R$$. For example, (due to the wikipedia article on polynomial rings) if $$R$$ is an integral domain, then so is $$R[x_1,...,x_n]$$ and if $$R$$ is a UFD, then so is $$R[x_1,...,x_n]$$.

General monoid rings are generalizations of the notion of a polynomial rings in finitely many variables. For example, if $$X$$ is an infinite set, then we can form a "polynomial ring" in $$|X|$$ many variables as $$R[M]$$ where $$M$$ is the free commutative monoid on $$X$$. Of course, if $$X = \{x_1,...,x_n\}$$, then this construction yields a usual polynomial ring $$R[x_1,...,x_n]$$. On the other hand, $$R[M]$$ is a free algebra on $$X$$ for the free (non-commutative) monoid $$M$$ on $$X$$.

My question is: what properties of $$R$$ are preserved by $$R[M]$$ for a monoid $$M$$? For instance, is being an integral domain is preserved? I would also like some references on the matter.

• Does an abelian monoid which is free on $\{x_i \mid i\in I\}$ necessarily isomorphic to $\oplus_{i\in I}\mathbb N$? I ask because I'm not completely certain: I don't know what dangers lie for intuition outside free abelian groups. If it is, then it seems like the same proof as for polynomial rings would hold. Feb 18, 2019 at 19:20
• @rschwieb: yes. Feb 18, 2019 at 22:04
• The only thing I can think of that is preserved is that if $M$ is finitely generated and abelian, then $R[M]$ is Noetherian iff $R$ is, by Hilbert's basis theorem.
– jgon
Feb 18, 2019 at 22:18

No, being an integral domain is not preserved; for example, if $$M$$ is the free idempotent (the free monoid on an element $$m$$ satisfying $$m^2 = m$$) then $$R[M] \cong R[m]/(m^2 - m) \cong R \times R$$. I don't know anything useful to say about what is preserved. The monoid ring construction is very general.
• What about the case when $M$ is the free monoid on an infinite set $X$ ($R[M]$ then being a polynomial ring in infinitely many variables)? Feb 18, 2019 at 23:15
• @Jxt921: assuming you mean "free commutative monoid," then as you say, $R[M]$ is a polynomial ring, and is an integral domain and a UFD if $R$ is (math.stackexchange.com/questions/1266784/…). Feb 19, 2019 at 23:52