I am given that $k$ is a field and $G$ is the monoid consisting of all monomials $X^iY^j$, where $j$ is between $0$ and $3i$. I am trying to compute the quotient of the monoid algebra $kG$ by the ideal generated by $X$ and the localization of $kG$ at the element $X$. I thought the quotient ring is isomorphic to $k$, since all monomials $X^iY^j$ where $i$ is positive collapse to zero in the quotient. Is this correct? How could I compute the localization of $kG$ at the element $X$? I was also thinking that we could visualize the ring $kG$ using the cone in $\mathbb{R}^2$ generated by $e_1$ and $e_1+3e_2$. Maybe this visualization would be helpful? Your help will be very much appreciated.
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$\begingroup$ $X$ and $Y$ commute in your monoid? $\endgroup$– LSpiceCommented Apr 21, 2022 at 15:45
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1$\begingroup$ Thanks for your question. Yes, X and Y commute in the monoid G. $\endgroup$– BorisCommented Apr 21, 2022 at 16:05
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$\begingroup$ Must $j$ be a multiple of $3$? That is, is $G$ supposed to be the monoid generated by the submonoid of the free commutative monoid on two generators generated by (1,0) and (1,3)? I think you might want to be more careful with your notation here because $XY^3$ is not a multiple of $X$ in this monoid. Your choice of using the embedding into the polynomial ring on X,Y makes things confusing. $\endgroup$– Benjamin SteinbergCommented Apr 21, 2022 at 20:40
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$\begingroup$ It seems to me that $(1,0)$, $(1,3)$ freely generate a free commutative monoid on two generators because you can check the second coordinate to know how many $(1,3)$'s were used. In that case $KG/(X)$ is just a polynomial ring in one variable. Note you should really not use G for a monoid but rather M. Similarly, if you view $KG$ as a polynomial ring in two variables one variable corresponding to $(1,0)$ and the other to $(1,3)$, then the localization question becomes much easier. $\endgroup$– Benjamin SteinbergCommented Apr 21, 2022 at 20:45
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$\begingroup$ Hello Benjamin. Thank you very much for your help and comments. J is only assumed to be between 0 and 3 times of I. So J need not be a multiple of 3. $\endgroup$– BorisCommented Apr 21, 2022 at 21:27
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