I am computing some divisors on a curve over a dvr and have run into some issues, which I would much appreciate clarification on. Specifically, I am considering the curve $\mathcal C \subset \mathbb P^2_{\mathbb Z_p}$ given by the projective equation $x^2 + p^2y^2 = z^2$ with $p \neq 2$. The special fibre of this is given by $x^2 = z^2$ and thus consists of two curves isomorphic to $\mathbb P^1_{\mathbb F_p}$ meeting at a point given by coordinates $(0,1,0)$ in $\mathbb P_{\mathbb F_p}^2$.
Considering one of these, given by $x=z$, gives me a divisor $D$. The ideal sheaf of $D$ on $\mathcal C$ should be invertible as $\mathcal C$ is nonsingular, and is given by $\mathcal I = (p,x-z)$. The pullback to $D$ gives the conormal sheaf $C_{D/\mathcal C}$, which should also be invertible.
However, when I attempt to calculate this sheaf, it appears to not be invertible. On the open set given by $y=1$, it is generated $p$ and $x-z$, with the latter satisfying the equation $(x+z)(x-z) = 2x(x-z) = 0$. This appears to give a rank 2 stalk at the point $x=z=0$ in the special fibre: where am I going wrong?