# prime divisor propertyfor Hurwitz integers

The Hurwitz integers $\mathcal{H}_{\mathbb{Z}}$ is a particular subset of quaternions. Define: $$\mathcal{H}_{\mathbb{Z}} = \left\{ a\frac{1+i+j+k}{2}+bi+cj+dk \ | \ a,b,c,d \in \mathbb{Z} \right\} = \mathbb{Z} \left[\frac{1+i+j+k}{2},i,j,k\right].$$ We say $\delta$ is a right divisor of $\alpha$ if there exists $\gamma$ for which $\alpha = \gamma \delta$. On the other hand (literally), we say $\delta$ is a left divisor of $\alpha$ if there exists $\gamma$ for which $\alpha = \delta \gamma$. In Stillwell's Elements of Number Theory page 152 the following weakened version of the prime divisor property is given:

Weak Prime Divisor Property: If $p$ is a real prime and if $p$ divides a Hurwitz integer product $\alpha \beta$, then $p$ divides $\alpha$ or $p$ divides $\beta$.

Here it is assumed $p \in \mathbb{Z}$ and $\alpha, \beta \in \mathcal{H}_{\mathbb{Z}}$. I am curious if the full prime divisor property:

Prime Divisor Property: If $p$ is a Hurwitz prime and if $p$ divides a Hurwitz integer product $\alpha \beta$, then $p$ divides $\alpha$ or $p$ divides $\beta$.

where $p,\alpha, \beta \in \mathcal{H}_{\mathbb{Z}}$. When I worked through a sketch of the proof of Stillwell's weakened prime divisor property I found I needed to use both left and right divisibility of $p$. I suspect, in general, left divisibility need not imply right divisibility and vice-versa. Thus, I fear the full prime divisor property is false for $\mathcal{H}_{\mathbb{Z}}$. Hence my question:

Question: Is the prime divisor property false for $\mathcal{H}_{\mathbb{Z}}$? Can you give an explicit counter-example. Or, alternatively, is it true and can you either provide or reference a proof?

I hope the third option; your question doesn't make sense is not a viable option. In any event, I thank the MSE in advance for its collective wisdom.

Sorry to disappoint, but I'm not so sure that even the "Weak Prime Divisor Property" you've stated actually holds for the Hurwitz Integers. For example, let $p=2$. Then $p$ factors as $p = (1+i)(1-i)$. Although $p$ divides the product, it divides neither of these two factors. Trust me, I so badly wish that this were true, but unfortunately the number theory of the Hurwitz integers is a bit harder to work with than that of the natural numbers.

The Hurwitz Integers do have properties that are useful for making divisibility arguments. They have a version of unique factorization, except that there are conditions you must be careful to consider when using it. This paper is a good source of theorems on this topic. I highly recommend it:

http://m-hikari.com/imf/imf-2012/41-44-2012/perngIMF41-44-2012.pdf

Another good source is Conway and Smith's book On Quaternions and Octonions.

An easy example is that $p=3$ factors as $p=(1+i+j)(1−i−j)$, yet 3 divides neither factor. The issue is that every natural number can be expressed as a sum of four squares, thus every real prime $p$ admits a factorization as $p=(a+bi+cj+dk)(a−bi−cj−dk)$. If $p$ were to divide either of these factors, it would have to divide the other as well (since in general $a+bi+cj+dk$ is divisible by a rational integer $n$ if $n$ divides each of $a,b,c$ and $d$). This would mean that the right side would have a factor of $p^2$, a contradiction.

• I suppose I forgot to say $p$ was an odd real prime for the weak case. I'll accept your answer if you can give an example of the prime divisor property failing for the non-real case. I think I have an example, but, I haven't proved all the details. Thanks for the link. – James S. Cook May 28 '15 at 13:25
• An easy example is that $p = 3$ factors as $p = (1 + i + j)(1 - i - j)$, yet $3$ divides neither factor. The issue is that every natural number can be expressed as a sum of four squares, thus every real prime $p$ admits a factorization as $p = (a+bi+cj+dk)(a-bi-cj-dk)$. If $p$ were to divide either of these factors, it would have to divide the other as well (since in general $a+bi+cj+dk$ is divisible by a rational integer $n$ iff $n$ divides each of $a,b,c$ and $d$). This would mean that the right side would have a factor of $p^2$, a contradiction. – KidA424 May 28 '15 at 21:20
• add your last comment to your answer and I'll accept it. Thanks! I'm new to this stuff and certain things are still sinking in. Very nice. – James S. Cook May 28 '15 at 21:55
• You're not alone...just keep playing around with examples. The more examples you see, the more you will internalize the theorems. That's what works for me, at least. – KidA424 May 28 '15 at 22:30