MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathbb Z_n$ denote the integers modulo $n$. Let $\mathbb Z_n[i, j, k]$ be the quaternionic ring over $\mathbb Z_n$, that is, the free module over $\mathbb Z_n$ with basis $\{1, i, j, k\}$ and multiplication defined by $$i^2=j^2=k^2=ijk=-1.$$

It is well-known that if $n=p$ where $p$ is a odd prime then $\mathbb Z_p[i, j, k]$ is isomorphic to the full matrix ring of order $2$ over $\mathbb Z_p$. What can be said about the structure of $\mathbb Z_n[i, j, k]$ for a composite $n$. Is also a full matrix ring?

Note:If we can prove that $\mathbb Z_n[i, j, k]$ is semi-simple then it is necessarily a direct sum of matrix rings over a field by a theorem of Wedderburn.

share|cite|improve this question
up vote 3 down vote accepted

Yes, for odd $n$, the ring $(\mathbb Z/n)[i,j,k]$ is isomorphic to the ring of two-by-two matrices over $\mathbb Z/n$.

To explain this, it is better to write $\mathbb Z/n$ and $\mathbb Z/p$ rather than $\mathbb Z_n$ and $\mathbb Z_p$, because, in fact, the decisive statement concerns $p$-adic integers $\mathbb Z_p$, and we should reserve the notation for that.

What is true is that, using the notation for $p$-adic integers, for odd prime $p$, $\mathbb Z_p[i,j,k]$ is the full matrix ring over the $p$-adic integers $\mathbb Z_p$. Granting this for a moment, $(\mathbb Z/p^\ell)[i,j,k]$ is the image of $\mathbb Z_p[i,j,k]$ by mapping $p^\ell \mathbb Z_p$ to $0$, since $\mathbb Z_p/p^\ell \mathbb Z_p\approx \mathbb Z/p^\ell \mathbb Z$.

By Sun-Ze's theorem, the general $\mathbb Z/n$ is the sum of the $\mathbb Z/p^\ell$ where $p^\ell$ are the prime powers dividing $n$. Thus, for odd $n$, $(\mathbb Z/n)[i,j,k]$ is the sum of the corresponding rings for the $p^\ell$.

To prove that $\mathbb Z_p[i,j,k]$ is the full matrix ring, start with the point that $\mathbb Q_p[i,j,k]$ is the full matrix ring, for odd $p$. There are various proofs of this... For example, the surjectivity of norms on finite fields, together with Hensel's Lemma, proves that the quadratic form $a^2+b^2+c^2+d^2=\hbox{norm}(a+bi+cj+dk)$ has a non-trivial zero, so $\mathbb Q_p[i,j,k]$ is not a division ring.

The additional ingredient is that the ring of "Hurwitz integers" consisting of $\mathbb Z[i,j,k]$ with $(1+i+j+k)/2$ adjoined is Euclidean, in the appropriate sense for a non-commutative ring. "Locally" at odd primes $p$, the $1/2$ in the definition of the Hurwitz integers is a unit, so, locally, at odd primes, the $p$-adic version of the Hurwitz integers is the naive notion of local quaternion integers, $\mathbb Z_p[i,j,k]$. The rest of the discussion is just clean-up.

share|cite|improve this answer
In addition, note that if $\mathbb H$ denotes the Hurwitz integers then $\mathbb H\otimes \mathbb Q_2$ is a field of quaternions (the underlying quadratic form being anisotropic). Thus $\mathbb H\otimes \mathbb Z/2^k$ will be anisotropic for $k$ big enough ($k>3$ should do the job) and won't be a subring of the ring of two by two matrices over $\mathbb Z/2^k$. – G.C. May 24 '12 at 23:58
thanks for your answer, is very useful for me. – miguel May 25 '12 at 13:32

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.