Timeline for How to compute the quotient and localization of the monoid algebra $kG$ for a field $k$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2022 at 23:38 | comment | added | Boris | Thanks for telling me this information. | |
| Apr 21, 2022 at 23:04 | comment | added | Benjamin Steinberg | That is different than generating with e1 and e1+3e2. That's why I was confused | |
| Apr 21, 2022 at 21:27 | comment | added | Boris | 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. | |
| Apr 21, 2022 at 20:45 | comment | added | Benjamin Steinberg | 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. | |
| Apr 21, 2022 at 20:40 | comment | added | Benjamin Steinberg | 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. | |
| Apr 21, 2022 at 16:05 | comment | added | Boris | Thanks for your question. Yes, X and Y commute in the monoid G. | |
| Apr 21, 2022 at 15:45 | comment | added | LSpice | $X$ and $Y$ commute in your monoid? | |
| Apr 21, 2022 at 15:37 | history | asked | Boris | CC BY-SA 4.0 |