# 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?

-
I'm trying to prove it for a ring $R$ such that the equation $x^2+y^2=-1$ has solutions with $x\neq 0$ and $y\neq 0$. – zacarias Jul 9 '12 at 21:20

(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.

-
It's a standard way of defining the multiplication. We get $ij=k$ by taking $ijk=-1$ and multiplying by $-k$ on the right. We get $ji=-k$ by multiplying $ijk=-1$ by $ji$ on the left, so we get $jiijk=-ji$, and on the left we get $j(-1)jk = -jjk = k$; so $k=-ji$, hence $ji=-k$. Then $ik=i(ij) = -j$, $ki=-jii = j$; multiplying $ijk=-1$ by $i$ on th eleft we get $-jk=-i$ or $jk=i$, and $kj = j(-ji) = i$. – Arturo Magidin Jul 10 '12 at 2:07
@Arturo: thanks. I did believe that those relations implied "mine". My point though is that it doesn't really make sense to define an algebra as the free $R$-module with a certain basis and then give relations that are anything other than the structure constants with respect to the basis. – Pete L. Clark Jul 10 '12 at 2:11
Maybe slightly strange, but $i^2 = j^2 = k^2 = ijk = -1$ are the relations that Hamilton carved into Brougham bridge. – t.b. Jul 10 '12 at 9:59
@t.b.: True, but Hamilton was not working over an arbitrary commutative ring (in which an $R$-module is not necessarily free). The point I'm trying to make is somewhat subtle and probably not that important: with the definition the OP gave, there's something to check to see that it even makes sense. – Pete L. Clark Jul 10 '12 at 10:21
@Pete: I agree with your main point. If I gave the derivation, it was mainly because you had expressed only that "maybe" they implied the relations as you presented them; it seemed that you hadn't checked or couldn't quite remember how to prove it, so I provided the derivation. But, again, I agree that it makes far more sense to define them via the structure constants. – Arturo Magidin Jul 10 '12 at 23:27