# 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. Commented Feb 18, 2019 at 19:20
• @rschwieb: yes. Commented 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
Commented 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)? Commented 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/…). Commented Feb 19, 2019 at 23:52
Sorry to comment on a very old question, but I wanted to add some information on the Cohen-Macaulay property for monoid rings. If you have never studied Cohen-Macaulayness of rings, Bruns and Herzog have an excellent book devoted to the subject. Melvin Hochster studied the Cohen-Macaulay property of monoid rings $$R[M]$$ in his paper "Rings of Invariants of Tori, Cohen-Macaulay Rings Generated by Monomials, and Polytopes", in the Annals of Mathematics, Sep. 1972.
What he showed is, modulo some details, if $$M$$ is a finitely generated sub-monoid of $$\mathbb{Z}^d$$ and $$k[M]$$ is integrally closed in its field of fractions for some field $$k$$, then $$R[M]$$ is Cohen-Macaulay for any (Noetherian) Cohen-Macaualay ring $$R$$. He applied this to show that some rings of invariants were Cohen-Macaulay.