Real Quaternions as a Division Ring

Detailed Solution

For a quaternion $$ q=a+bi+cj+dk, $$ define its conjugate by $$ \overline q=a-bi-cj-dk. $$ Then $$ q\overline q=\overline q q=a^2+b^2+c^2+d^2=N(q). $$ If $q\neq0$, then $$ N(q)=a^2+b^2+c^2+d^2>0. $$ Hence $$ q^{-1}=\frac{\overline q}{N(q)} $$ exists in $\mathbb H$. Therefore every nonzero quaternion is invertible, so $\mathbb H$ is a division ring. Moreover, $\mathbb H$ is not commutative because $$ ij=k,\qquad ji=-k, $$ and hence $ij\neq ji$. Thus $\mathbb H$ is a division ring but not a field. It also has no zero divisors. Indeed, if $pq=0$, then using multiplicativity of the norm, $$ N(pq)=N(p)N(q)=0. $$ Since $N(p),N(q)\ge0$, this implies $N(p)=0$ or $N(q)=0$, and hence $p=0$ or $q=0$. Therefore the correct option is $$ \boxed{\mathrm{B}}. $$