{\mathbb{Q}}[i]$ and let $D$ be its discriminant. If $R$ is the ring of integers of ${\mathbb{Q}}[i]$, then $R^\perp = Rj$ and $R + Rj$ is a suborder of index $|D|$ in ${\mathcal{O}}$. If $\omega$ is a generator of $R$, then $${\mathrm{Nrd}}(x_1 + y_1\omega + (x_2 + y_2\omega)j) = f(x_1,y_1) + p f(x_2,y_2),$$ where $f(x,y)$ is a principal quadratic form of discriminant $D$.

The triviality of the trace of $j$ and the anti-commuting relation $ij = -ji$ imply that ${\mathbb{Q}}(i)$ has orthogonal complement ${\mathbb{Q}}(i)j$ in $B_{p,\infty}$. Consequently $R^\perp \subset {\mathcal{O}}$ is a lattice in ${\mathbb{Q}}(i)j$ containing $Rj$, hence of the form $\mathfrak{a}j$ for a fractional ideal $\mathfrak{a}$ of $R$ which contains $R$. The prime $p$ is inert in $R$, since $p$ is ramified in $B_{p,\infty}$ but not in $R$. Since the norm is integral on $\mathfrak{a}j$, and ${\mathrm{Nrd}}(j) = p$, it follows that $\mathfrak{a}$ is integral, hence equals $R$. The orthogonality of $R$ and $Rj$ implies that $j\beta = \bar{\beta}j$ for all $\beta$ in $R$, so $jR = Rj$ and $R + Rj$ is closed under multiplication. The form of the norm follows from orthogonality and multiplicativity of the norm: ${\mathrm{Nrd}}(\beta_1 + \beta_2j) 
= {\mathrm{Nrd}}(\beta_1) + p {\mathrm{Nrd}}(\beta_2)$. Consequently the discriminant 