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 $a\_0, a\_1, a\_2, a\_3\in\mathbb Z\_{2^s}$.



Rings of Quaternions 
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 $s$.


