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This is elementary, but I have no idea how to solve it for all $s$. Actually I'm not sure if it's true. Let $\mathbb Z_{2^s}$ be the ring of integers modulo $2^s$. Considere the hamilton quaternions over $\mathbb Z_{2^s}$, that is, the ring

$\mathbb H_{2^s}=\mathbb Z_{2^s}[i, j, k]$={$a_0+a_1i+a_2j+a_3k\;\;:\;\;a_l\in\mathbb Z_{2^s}$},

where each element of $\mathbb Z_{2^s}$ commute with $i, j$ and $k$, and $i^2=j^2=k^2=ijk=-1$.

Is $\mathbb H_{2^s}$ a semicommutative ring? that is, for every $z\in\mathbb H_{2^s}$ the right annihilator $ann_r(z)$ is an ideal of $\mathbb H_{2^s}$.

Note that reversible rings (rings where $xy=0$ implies that $yx=0$) are semicommutative. My intuition is that $\mathbb H_{2^s}$ is a reversible ring.

It can help to observe that $\mathbb H_{2^s}$ is a local ring and $\mathbb H_{2^s}/J(\mathbb H_{2^s})\cong\mathbb Z_2 $, where $J$ denotes de jacobson radical. I also have simple proofs that $J(\mathbb H_{2^s})$ is generated by $1+i$, $ 1+j$ and $1+k$ and every element $z$ of $J(\mathbb H_{2^s})$ can be written as $z=2a_0+a_1(1+i)+a_2(1+j)+a_3(1+k)$, where $a_0, a_1, a_2, a_3\in\mathbb Z_{2^s}$.

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up vote 4 down vote accepted

Assume $xy=0$ but $yx\neq 0$. We can take $x$ and $y$ to be $2$-adic quaternions, and then we have $xy=0$ modulo $2^s$ but $yx\neq 0$ modulo $2^s$. We can factor out the highest possible power of $2$ from $x$ and $y$, and then divide $2^s$ by that power, so without loss of generality $x$ and $y$ are nontrivial modulo $2$.

$xy\bar{y}=x|y|^2$. $|y|^2$ is a regular $2$-adic number and is the sum of four squares, at least one odd, and so is nonzero mod $8$. $x$ has a coefficient that is nonzero mod $2$ so that same coefficient in $x|y|^2$ is nonzero mod $8$, so $x|y|^2$ is nonzero mod $8$, so $xy$ is nonzero mod $8$.

Therefore $s$ must be $1$ or $2$. Since a-fortiori computed that reversibility holds for $s=1,2$, it holds for every $s$.

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The OP is not talking about $\mathbb{Z}_2$ but about $\mathbb{Z}/2^s \mathbb{Z}$. –  Qiaochu Yuan Jun 25 '12 at 6:25
    
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. –  zacarias Jun 25 '12 at 11:34
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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 $2^s$. –  David Speyer Jun 25 '12 at 12:48
    
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 $\mathbb H_{2^s}$, $s=1, 2, 3, \ldots$. –  zacarias Jun 25 '12 at 13:20
    
I get that product to be $4j$. In noncommutative rings, it's not true that $a^2-b^2 = (a-b)(a+b)$. –  Konstantin Ardakov Jun 25 '12 at 14:48
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