Timeline for Group rings of free abelian groups
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 9, 2021 at 13:34 | comment | added | YCor | The Krull dimension of $\mathbf{Z}[\mathbf{Z}^d]$ is $d+1$. And also for every torsion-free abelian group $G$, the group of units of $\mathbf{Z}[G]/2\mathbf{Z}[G]$ is reduced to $G$ (as mentioned by Benjamin Steinberg, this extends to the case when $G$ is left-orderable). | |
| Dec 9, 2021 at 13:33 | history | edited | Martin Sleziak |
added the (group-rings) tag
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| Dec 9, 2021 at 13:32 | comment | added | Benjamin Steinberg | Sorry, for free abelian this is easier but it is true for any abelian groups. Free abelian groups are left orderable and so satisfy the Kaplansky unit conjecture. You can find in Passman's group ring book that if $G_1$ satisfies the Kaplansky unit conjecture and $\mathbb ZG_1\cong \mathbb ZH$ then $G_1\cong H$. | |
| Dec 9, 2021 at 13:30 | answer | added | Benjamin Steinberg | timeline score: 10 | |
| Dec 9, 2021 at 13:23 | comment | added | Benjamin Steinberg | It is corollary 3 here ams.org/journals/tran/1969-136-00/S0002-9947-1969-0233903-9/… | |
| Dec 9, 2021 at 9:21 | comment | added | Jeremy Rickard | @NarutakaOZAWA I think you want to count homomorphisms to $\mathbb{Z}/3\mathbb{Z}$ rather than to $\mathbb{Z}/2\mathbb{Z}$. For any $G$ there's only one ring homomorphism $\mathbb{Z}[G]\to\mathbb{Z}/2\mathbb{Z}$. | |
| Dec 9, 2021 at 3:56 | comment | added | Zach Teitler | Hopefully this earlier answer is helpful: mathoverflow.net/a/299923 | |
| Dec 9, 2021 at 3:17 | comment | added | Narutaka OZAWA | I don't know a reference, but can count $\operatorname{Hom}(\mathbb{Z}[G],\mathbb{Z}/2\mathbb{Z})$. | |
| S Dec 9, 2021 at 0:29 | review | First questions | |||
| Dec 9, 2021 at 2:57 | |||||
| S Dec 9, 2021 at 0:29 | history | asked | Ivan Degtyar | CC BY-SA 4.0 |