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.

  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Feb 18, 2019 at 19:20
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    $\begingroup$ @rschwieb: yes. $\endgroup$ Feb 18, 2019 at 22:04
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    $\begingroup$ 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. $\endgroup$
    – jgon
    Feb 18, 2019 at 22:18

1 Answer 1


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.

  • $\begingroup$ 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)? $\endgroup$
    – Jxt921
    Feb 18, 2019 at 23:15
  • $\begingroup$ @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/…). $\endgroup$ Feb 19, 2019 at 23:52

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