Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.