Showing that an element is a unit in a ring

[Mr Davis 97]

Homework Statement

Let $S=\{a+bi+cj+dk \mid a,b,c,d \in \mathbb{Z}\}$ be the ring of integral Hamiltonian quaternions, where multiplication is defined using the same rules as in $\mathbb{H}$, the ring of real Hamiltonian quaternions. Define a function $$N:S\to\mathbb{Z}, N(a+bi+cj+dk)=a^2+b^2+c^2+d^2.$$

Show that if $N(\alpha)=1$, then $\alpha$ is a unit

Homework Equations

The Attempt at a Solution

So suppose we know that $N(\alpha) = \alpha\bar{\alpha}$, where the bar is the conjugate of the quaternion. If $N(\alpha)=1$, then $\alpha\bar{\alpha} = 1$. So to show that $\alpha$ is a unit, all I need to show is the other direction, that $\bar{\alpha}\alpha=1$. How can I do this? In groups one implies the other since we have inverses, but this is not the case with rings.

 

[fresh_42]

If you already know the inverse, show that it is in $S$ and that it actually is the inverse.

 

[Mr Davis 97]

If you already know the inverse, show that it is in $S$ and that it actually is the inverse.

Well all I know is that $\alpha\bar{\alpha} = 1$. Clearly $\bar{\alpha}\in S$, so all I need to show that $\bar{\alpha}\alpha = 1$. But I don't see how to get that

 

[Dick]

Well all I know is that $\alpha\bar{\alpha} = 1$. Clearly $\bar{\alpha}\in S$, so all I need to show that $\bar{\alpha}\alpha = 1$. But I don't see how to get that

If $\alpha=a+bi+cj+dk$ and you define $\bar{\alpha}=a-bi-cj-dk$ so that $\alpha \bar \alpha=a^2+b^2+c^2+d^2$, doesn't that make $\bar \alpha \alpha=a^2+(-b)^2+(-c)^2+(-d)^2$?