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I am looking for examples of monoid rings $R(M)$ that is a quotient $R[X_1,\ldots]/{\mathfrak a}$ of a polynomial ring $R[X_1,\ldots ]$ which any number of indeterminates and ${\mathfrak a}$ is a homogeneous ideals.

I have two examples. First is $R({\mathbb N})$ where ${\mathbb N}$ is the natural numbers under addition. Then $R({\mathbb N})\cong R[X]$.

Let $p$ be a prime. Another example is ${\mathbb F}_p({\mathbb Z}/p)$ where ${\mathbb F}_p$ is the finite field with $p$ elements and ${\mathbb Z}/p$ is the cyclic group of integers mod $p$. Then $${\mathbb F}_p({\mathbb Z}/p) \cong {\mathbb F}_p[\bar{t}]/(\bar{t}^p)$$ where $t$ is a generator of ${\mathbb F}_p$ and $\bar{t}=t-1$.

Do you have other examples or general sufficient conditions?

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  • $\begingroup$ Another example: Laurent polynomials in one variable, obtained as the quotient of the polynomial ring on two generators $x$ and $y$ by the relation $xy=1$. $\endgroup$
    – Yemon Choi
    Oct 19, 2012 at 8:14
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    $\begingroup$ $xy=1$ doesn't look homogeneous in $R[x,y]$. $\endgroup$ Oct 19, 2012 at 13:45
  • $\begingroup$ If by $R[X, X^{-1}]$ you mean a polynomial ring with two indeterminates, then that is not isomorphic to the ring of Laurent polynomials. $\endgroup$
    – user2529
    Oct 26, 2012 at 6:19
  • $\begingroup$ By $A=R[X^{\pm 1}]$ I mean the ring $R[X,Y]/(XY-1)$. Clearly, $A$ is the ring of Laurent polynomials and it is also isomorphic to the group ring $R[\mathbb Z]$. There are other ways to see this ring that are connected with the view-point of monoid rings. In fact, if you take the monoid ring $R[\mathbb N]\cong R[X]$, and you consider the multiplicative system $\Sigma_X=${$X,X^2,\dots,X^n,\dots$} of central elements in $\Sigma$, then the ring $R[\mathbb Z]\cong R[X^{\pm 1}]$ is the Ore localization $\Sigma^{-1}R[X]$. $\endgroup$ Oct 28, 2012 at 23:06
  • $\begingroup$ On a second look, it still seems that the ring $R[X,Y]/(XY-1)$ of Laurent examples doesn't answer my question since $XY-1$ is not homogeneous. (to repeat Fernando.) Another thing, I have realized that if a group monoid $R(M)$ is a quotient of a polynomial ring, then necessarily $M$ is abelian. $\endgroup$
    – user2529
    Nov 13, 2012 at 14:09

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