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Problem: Consider the Quaternions over a general commutative ring $R$ instead of $\Bbb R$, say $H(R)$.

I want to show that $q \in H(R)$ is a unit iff $N(q)$ is a unit in $R$. If $q=a+bi+cj+dk$, then $N(q) = a^2+b^2+c^2+d^2$. Let $q^*$ be the conjugate of $q$, aka $q^* = a -bi -cj -dk$.

Attempt

If $N(q)$ is invertible in $R$, then $\frac {q^*} {qq^*} = \frac {q^*} {N(q)}$ is an inverse of $q$, so that $q$ is a unit.

The other direction is giving me trouble. If $q$ is invertible, is it necessarily true that its inverse must be of the form $\frac {q^*} {N(q)}$ ?

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The point is that $N$ is multiplicative: $N(qq')=N(q)N(q')$. If $q$ is invertible, say $qq'=1$, then $N(q)N(q')=N(1)=1$ and so $N(q)$ is a unit.

Then $N(q)=qq^*=q^*q$ and so $q(q^*/N(q))=(q^*/N(q))q=1$ etc.

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    $\begingroup$ Ok. I need to work out that $N(qq')=N(q)N(q')$ for myself but thanks! $\endgroup$ – IntegrateThis Jan 21 at 19:39

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