Timeline for Rings of Quaternions
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 25, 2012 at 20:00 | vote | accept | zacarias | ||
| Jun 25, 2012 at 20:00 | comment | added | zacarias | Dear Will, thank you, I think your proof works. | |
| Jun 25, 2012 at 15:36 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jun 25, 2012 at 15:34 | comment | added | zacarias | dear Will, I think your example does not work, Konstantin is right. In fact a ring $R$ is semicommutative is and only if $ab=0$ implies that $aRb=0$ , for all $a, b\in R$ . If $a=1-i-j-k$ and $b=1+i+j+k$ are in $\mathbb H_{2^2}$ (as in your counterexample) we have $ab=0$ , but also $aib=0$ so we can not conclude that $a\mathbb H_{2^2}b\neq 0$ . can you explain why if there is a counterexample for some $s$ there is necessarily a couterexample for $s=2$? This solves the problem since $\mathbb H_{2^2}$ is fine we can test all possible cases. | |
| Jun 25, 2012 at 15:31 | comment | added | Will Sawin | Alright. Based on a-fortiori's comment I'll give that argument. | |
| Jun 25, 2012 at 15:01 | comment | added | user2035 | Computer says that the ring is reversible at least for $s\leq 4$. | |
| Jun 25, 2012 at 14:48 | comment | added | user91132 | I get that product to be $4j$. In noncommutative rings, it's not true that $a^2-b^2 = (a-b)(a+b)$. | |
| Jun 25, 2012 at 14:14 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| Jun 25, 2012 at 13:20 | comment | added | zacarias | hello David Speyer, probably you're right, since over the $2$-adics the Hamiltonian quaternions have no zero divisors and thus reversibility is trivial. Probably this can help since $\mathbb H_{2^s}$, for every $s$, is an homomorphic image of the hamiltonian queternions over the $2$-adics. Also the quaternions over the $2$-adics is the inverse limite of $\mathbb H_{2^s}$, $s=1, 2, 3, \ldots$. | |
| Jun 25, 2012 at 12:48 | comment | added | David E Speyer | I am fairly sure that Will is using $\mathbb{Z}_2$ to mean the $2$-adics, not the integers modulo $2$. But, as Qiaochu says, the actual problem is about the integers modulo $2^s$. | |
| Jun 25, 2012 at 11:34 | comment | added | zacarias | thank you Will, but The problem (as stated by Qiaochu) is not only for $\mathbb Z_2$ but for $\mathbb Z_{2^s}$, for a positive integer $s$. For $s=1$ the ring $\mathbb H_2$ is commutative so this case is trivial. | |
| Jun 25, 2012 at 6:25 | comment | added | Qiaochu Yuan | The OP is not talking about $\mathbb{Z}_2$ but about $\mathbb{Z}/2^s \mathbb{Z}$. | |
| Jun 25, 2012 at 4:47 | history | answered | Will Sawin | CC BY-SA 3.0 |