Quaternion Rings Let $R$ be a commutative ring. Define the Hamilton quaternions $H(R)$ over $R$ to be the free $R$-module with basis $\{1, i, j, k\}$, that is,
$$H(R)=\{a_0+a_1i+a_2j+a_3k\;\;:\;\;a_l\in R\}.$$
and multiplication is defined by: $i^2=j^2=k^2=ijk=-1$.
Is well-known that over a field $F$ (with char $F\neq 2$) the ring $H(F)$ is a division ring or isomorphic to $M_2(F)$. What can we say  about the Hamilton quaternions over an arbitrary commutative ring $R$? Is still true that $H(R)$ is a division ring or isomorphic to $M_2(R)$? We must impose some conditions to the ring to make it happen?
 A: (Your definition of the multiplication looks slightly strange to me.  I would prefer to write $k = ij = -ji$.  This implies your relations; as Arturo shows in his comment below, your relations also imply mine, but to my mind it's sort of mixing two different ways to give an $R$-algebra: you could either give it via generators and relations, or if it's a free $R$-module you can give an explicit basis and the structure constants.)
If $R$ is a field of characteristic different from $2$, then $H(R)$ is by definition the quaternion algebra $\left( \frac{-1,-1}{R} \right)$.  In general, for $a,b \in R^{\times}$, the quaternion algebra
$\left( \frac{a,b}{R} \right) = R\langle i,j \rangle/(i^2 = a, j^2 = b, ji = -ij)$
is indeed always a division algebra or isomorphic to $M_2(R)$: see e.g. $\S 5.1$ of these notes.
(If $R$ is a field of characteristic $2$, then $H(R)$ is a commutative ring, so is not what you want.  There are analogues of quaternion algebras in characteristic $2$ defined slightly differently.)
If $R$ is a domain, then the center of $H(R)$ is $R$.  It follows that if $R$ is not a field, $H(R)$ is not a division ring.  Note also that if $R \subset S$ is an extension of domains, then $H(S) \cong H(R) \otimes_R S$.
In particular $H(\mathbb{Z})$ is not a division ring.  Neither is it isomorphic to $M_2(\mathbb{Z})$: if so, then $H(\mathbb{\mathbb{R}})$ would be isomorphic to $M_2(\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R} \cong M_2(\mathbb{R})$, but $H(\mathbb{R}) = \left( \frac{-1,-1}{\mathbb{R}}\right)$ is the classical Hamiltonian quaternions, which is well-known to be a division ring.
In general, if $R$ is a domain of characteristic different from $2$ with fraction field $K$, then $H(R)$ is an order in the quaternion algebra $H(K)$.  This is more a definition than anything else, but it gives you a name.  There is a vast literature on quaternion orders: for just a little bit, see e.g. these notes.
