I'm working with Hales The Group Law of Edwards Curves where he defines the addition $z_1 \oplus z_2 = (\frac{x_1x_2 - c y_1y_2}{1 - d x_1x_2y_1y_2}, \frac{x_1y_2+y_1x_2}{1+dx_1x_2y_1y_2})$ and he writes this is an element of $R_2[\frac 1 \delta] \times R_2[\frac 1 \delta]$ where $R_2[\frac 1 \delta]$ is the localization of $R_2 = \mathbb{Z}[c,d][x_1,y_1,x_2,y_2]$ with respect to the multiplicative set $S = \{1,\delta,\delta^2,\ldots\}$ where $\delta = \delta^- \delta^+ = (1 - d x_1x_2y_1y_2) (1+dx_1x_2y_1y_2)$
My problem is the case $\delta = 0$. I think with this notation the operation is still defined as $z_1 \oplus z_2 = ((x_1x_2 - c y_1y_2)(1+dx_1x_2y_1y_2),(x_1y_2+y_1x_2)(1 - d x_1x_2y_1y_2))$. This would mean I have to distinguish in my proofs the case $\delta = 0$ and $\delta \neq 0$.
Am I correct? Or can I safely assume that $\delta \neq 0$ in the proof of the group law of Edwards curves?