# Show that $q$ is a unit in $H(R)$ iff $N(q)$ is a unit in $R$

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 $$\frac {q^*} {N(q)}$$ ?

## 1 Answer

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.

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