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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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – James S. Cook May 28 '15 at 13:25
  • $\begingroup$ 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. $\endgroup$ – KidA424 May 28 '15 at 21:20
  • $\begingroup$ 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. $\endgroup$ – James S. Cook May 28 '15 at 21:55
  • $\begingroup$ 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. $\endgroup$ – KidA424 May 28 '15 at 22:30

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