Let $A$ be a commutative ring, $\alpha,\beta,\gamma\in A$ and $E$ the quaternion algebra over $A$ whose underlying module is $A^{\oplus 4}$ and whose multiplication table is given by \begin{align} i^2&=\alpha e+\beta i, & ij&=k, & ik&=\alpha j+\beta k,\\ ji&=\beta j-k, & j^2&=\gamma e, & jk&=\beta\gamma e-\gamma i,\\ ki&=-\alpha j, & kj&=\gamma i, & k^2&=-\alpha\gamma i, \end{align} where $(e,i,j,k)$ is the canonical basis of $A^{\oplus 4}$. Suppose $1+1\ne0$ in $A$. Then $E$ is not commutative.
Let $N_E$ denote the Cayley norm of $E$. Then $$N_E(\rho e+\xi i+\eta j+\zeta k)=(\rho^2+\beta\rho\xi-\alpha\xi^2-\gamma(\eta^2+\beta\eta\zeta-\alpha\zeta^2))e,$$ for all $\rho,\xi,\eta,\zeta\in A$. It can be shown that $N_E(uv)=N_E(u)N_E(v)$ for all $u,v\in E$. The author claims that this latter equality together with $1+1\ne0$ imply that $E$ is not commutative. I cannot see this. Any suggestions how one might prove this?