Timeline for Rings with group of units cyclic of prime order
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2013 at 21:18 | vote | accept | CommunityBot | ||
| Aug 5, 2013 at 20:30 | comment | added | Will Sawin | The same argument says that the unit group of a ring with an odd number of units must be a product of cyclic groups whose orders are Mersenne numbers. So if one requires the group to be cyclic of prime power order, say $p^n$, then one requires $p^n$ to be a Mersenne number. If it need not be cyclic, there are more possibilities. | |
| Aug 5, 2013 at 16:04 | comment | added | dimo | @WillSawin Nice answer, worth upvoting. Have any idea if we replace $p$ by $p^2$ or $p^n$ ?? | |
| Aug 5, 2013 at 16:02 | comment | added | dimo | @QiaochuYuan you are right. That was a misunderstanding. | |
| Aug 5, 2013 at 13:28 | comment | added | Qiaochu Yuan | @dimo: Will isn't claiming that either. The claim is that some direct factor of $\mathbb{F}_2[x](x^p - 1)$ has exactly $p$ invertible elements. | |
| Aug 5, 2013 at 8:13 | comment | added | dimo | @QiaochuYuan $R$ is a quotient of $\mathbb{F}_2[x]/(x^p - 1)$. How do you drive that $\mathbb{F}_2[x]/(x^p - 1)$ has exactly $p$ invertible elements ? | |
| Aug 4, 2013 at 21:53 | comment | added | Qiaochu Yuan | @Sally: Will isn't claiming that the kernel is $(x^p - 1)$. He only needs that the map is surjective, from which it follows that $R$ is a quotient of $\mathbb{F}_2[x]/(x^p - 1)$, and then he classifies all such quotients. | |
| Aug 4, 2013 at 20:26 | comment | added | Will Sawin | Send $x$ to a generator of the group of units of $R$. | |
| Aug 4, 2013 at 19:33 | comment | added | Todd Trimble | That's a nice analysis. | |
| Aug 4, 2013 at 19:25 | history | answered | Will Sawin | CC BY-SA 3.0 |