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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.

Dylan C. Beck
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Jxt921
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  • 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. – rschwieb Feb 18 '19 at 19:20
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    @rschwieb: yes. – Qiaochu Yuan Feb 18 '19 at 22:04
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    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 '19 at 22:18

2 Answers2

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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.

Qiaochu Yuan
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  • 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)? – Jxt921 Feb 18 '19 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 (https://math.stackexchange.com/questions/1266784/is-a-polynomial-ring-over-a-ufd-in-countably-many-variables-a-ufd). – Qiaochu Yuan Feb 19 '19 at 23:52
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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.

walkar
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