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Let $\mathcal{H}$ denote the Hurwitz Quaternion, i.e the subring of the ring of real quaternions that is defined in the following way: $$\mathcal{H} = \{ m_0 \zeta + m_1 i + m_2 j + m_3 k \mid m_i \in \mathbb{Z} \}$$ where $\zeta = \frac{1}{2}(1 + i + j + k)$. Let $N$ be the standard norm defined on the ring of real quaternions. i.e $N(x) = x_0^2 + x_1^2 + x_2^2 + x_3^2$ for $x = x_0 + x_1 i + x_2 j + x_3 k$

The claim is that norm of any non-zero Hurwitz Quaternion is always a positive integer. I am stuck on one piece of the argument. Here is what I have so far:

Let $x$ be an arbitrary element of $\mathcal{H}$ such that $x = m_0 \zeta + m_1 i + m_2 j + m_3 k$ , $m_i \in \mathbb{Z}$. Then we have that $$ x = \frac{m_0}{2} + \bigg{(} \frac{2m_1 + m_0}{2} \bigg{)} i + \bigg{(} \frac{2m_2 + m_0}{2} \bigg{)} j + \bigg{(} \frac{2m_3 + m_0}{2} \bigg{)} k $$ and so \begin{align*} \begin{split} N(x) &= \bigg{(} \frac{m_0}{2} \bigg{)}^2 + \bigg{(} \frac{2m_1 + m_0}{2} \bigg{)}^2 + \bigg{(} \frac{2m_2 + m_0}{2} \bigg{)}^2 + \bigg{(} \frac{2m_3 + m_0}{2} \bigg{)}^2\\\\ &= \frac{m_0^2 + 4m_1^2 + 4m_1m_0 + m_0^2 + 4m_2^2 + 4m_2m_0 + m_0^2 + 4m_3^2 + 4m_3m_0 + m_0^2}{4}\\\\ &= \frac{4(m_0^2 + m_1^2 + m_2^2 + m_3^2 + m_1m_0 + m_2m_0 + m_3m_0)}{4}\\\\ &= m_0^2 + m_1^2 + m_2^2 + m_3^2 + m_1m_0 + m_2m_0 + m_3m_0 \in \mathbb{Z} \end{split} \end{align*}

The last three terms are concerning me. Why can I argue that the sum of $m_1m_0, m_2m_0, m_3m_0$ are going to enable the norm to always be a positive integer?

Math 2tor
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An Isomorphic Teen
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1 Answers1

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As you wrote yourself, the norm is the sum of four squares and thus non-negative. If you now further assume that not all the $m_i$ are zero, we have thus a positive integer.

RobertMuller
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  • I agree that the sum of four squares is positive, I am asking as why this is clear after I teased out the norm for an arbitrary non-zero element of $\mathcal{H}$? – An Isomorphic Teen Dec 21 '22 at 14:57
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    I don't claim that it is clear after you expand the squares etc. Of course you can make some manipulations that make it harder to observe certain properties. The point is that you just don't need to deal with this, since you already know that it is positive (or in the general case at least non-negative) just by definition. There is just no reason to make all these manipulations for seeing that. Your computations still verifies that it is an integer and not some fraction though. – RobertMuller Dec 21 '22 at 15:01
  • I see what you are saying. thank you – An Isomorphic Teen Dec 21 '22 at 15:05