Timeline for Units in the group ring over fours group after Gardam
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 22, 2021 at 10:54 | comment | added | ARG | @Ycor I did not have any specific in mind (this is why I used the word "feeling"). If I have the time, maybe I'll think it through, but I have (unfortunately, because this is fun) more pressing things to attend. So just ignore this last comment. | |
| Mar 22, 2021 at 10:47 | comment | added | YCor | @ARG I have no idea what you mean by "the only zero-divisors are those of $K$"... if there are zero divisors in $K$, say $ab=0$, $a,b\neq 0$ and $G\neq\{1\}$, then there are plenty of zero divisors outside $K$ (such as $a\delta_g$ or $a(\delta_1-\delta_g)$ for $g\neq 1$. | |
| Mar 22, 2021 at 10:03 | comment | added | ARG | @Ycor still I have the feeling that one could improve my statement by saying that if the only zero-divisors are those of $K$ then the only idempotents are those of $K$ (but I had enough fun with the current argument). One question which still tickles me is whether one can do something with skew fields. | |
| Mar 22, 2021 at 10:00 | comment | added | ARG | @VilleSalo just when you think you have a nice argument, there comes a pundit to rob you of it (no offense to Ycor meant and no content added, just a silly [and perhaps overused] pun). | |
| Mar 22, 2021 at 9:28 | comment | added | Ville Salo | Actually the main point of my comment was probably to make the augmentation pun. | |
| Mar 22, 2021 at 9:26 | comment | added | Ville Salo | So we agree that ARG shows that "no zero divisors in $RG$ implies no torsion in $RG$" holds in all commutative rings (as I assume I meant with my comment), and you're saying it's silly to include non-domains, since the implication is trivial for them? | |
| Mar 22, 2021 at 9:20 | comment | added | YCor | @VilleSalo it's not really "all commutative rings" but rather "all domains". Indeed the zero divisor conjecture for fields implies it for domains, and conversely $RG$ has zero divisors when $R$ is not a domain. | |
| Mar 21, 2021 at 19:41 | history | edited | ARG | CC BY-SA 4.0 |
changed intro
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| Mar 15, 2021 at 8:47 | comment | added | ARG | @VilleSalo a bit of augmentation and a bit of Frobenius :D | |
| Mar 15, 2021 at 6:11 | comment | added | Ville Salo | This looks correct. So it seems your original argument only needed a little bit of augmentation to deal with all commutative rings. | |
| Mar 14, 2021 at 21:36 | history | answered | ARG | CC BY-SA 4.0 |