Let's try to take what we did when we were working with Z [ − d ] \mathbb{Z}[\sqrt{-d}] Z [ − d ]
Let's work with the quaternions . In the quaternions, let's define the conjugate of α = a + b i + c j + d k \alpha=a+bi+cj+dk α = a + bi + c j + d k α ‾ = a − b i − c j − d k \overline\alpha=a-bi-cj-dk α = a − bi − c j − d k N ( α ) = α α ‾ . N(\alpha)=\alpha\overline{\alpha}. N ( α ) = α α .
Right, we can say that it is multiplicative, using essentially the same proof.
N ( α β ) = α β α β ‾ = α α ‾ β β ‾ = N ( α ) N ( β ) . \begin{align}N(\alpha\beta) &= \alpha\beta\overline{\alpha\beta}\\
&= \alpha\overline{\alpha}\beta\overline\beta\\
&= N(\alpha)N(\beta).\end{align} N ( α β ) = α β α β = α α β β = N ( α ) N ( β ) .
Now, let's get messy. Let α = a + b i + c j + d k \alpha=a+bi+cj+dk α = a + bi + c j + d k β = w + x i + y j + z k \beta=w+xi+yj+zk β = w + x i + y j + z k
Let's use N ( α β ) = N ( α ) N ( β ) N(\alpha\beta)=N(\alpha)N(\beta) N ( α β ) = N ( α ) N ( β ) ( a 2 + b 2 + c 2 + d 2 ) ( w 2 + x 2 + y 2 + z 2 ) . (a^2+b^2+c^2+d^2)(w^2+x^2+y^2+z^2). ( a 2 + b 2 + c 2 + d 2 ) ( w 2 + x 2 + y 2 + z 2 ) .
Now let's find the left side. I'll do the expanding so that you don't.
( a w − b x − c y − d z ) + ( a x + b w + c z − d y ) i + ( a y − b z + c w + d x ) j + ( a z + b y − c x + d w ) k (aw-bx-cy-dz)+(ax+bw+cz-dy)i+(ay-bz+cw+dx)j+(az+by-cx+dw)k ( a w − b x − cy − d z ) + ( a x + b w + cz − d y ) i + ( a y − b z + c w + d x ) j + ( a z + b y − c x + d w ) k Great.
Now we take the norm, and we know that
( a 2 + b 2 + c 2 + d 2 ) ( w 2 + x 2 + y 2 + z 2 ) = ( a w − b x − c y − d z ) 2 + ( a x + b w + c z − d y ) 2 + ( a y − b z + c w + d x ) 2 + ( a z + b y − c x + d w ) 2 . \begin{align}(a^2+b^2+c^2+d^2)(w^2+x^2+y^2+z^2)=&(aw-bx-cy-dz)^2\\&+(ax+bw+cz-dy)^2\\&+(ay-bz+cw+dx)^2\\&+(az+by-cx+dw)^2. \end{align} ( a 2 + b 2 + c 2 + d 2 ) ( w 2 + x 2 + y 2 + z 2 ) = ( a w − b x − cy − d z ) 2 + ( a x + b w + cz − d y ) 2 + ( a y − b z + c w + d x ) 2 + ( a z + b y − c x + d w ) 2 .