Divisor class group of $\operatorname{Spec} k[x,y,z]/(xy-z^2)$ is $\mathbb{Z}/2 \mathbb{Z}$.

I'm trying to understand the following fact from Hartshorne. Let $A = k[x,y,z]/(xy-z^2)$ where $k$ is a field. Let $X = \operatorname{Spec} A$. Then the divisor class group of $X$, $\operatorname{Cl}(X) \cong \mathbb{Z}/ 2\mathbb{Z}$.

Let $Y = V(\overline{y}, \overline{z})$. Can anyone please explain the following facts in detail?

a) The ideal $(\overline{y}, \overline{z})$ is prime in $k[x,y,z]/(xy-z^2)$.

b) $X - Y $ is $\operatorname{Spec} (k[x,y,y^{-1}, z]/(xy - z^2)) $.

c) Divisor of $\overline{y}$ is equal to $2Y$. How is $v_{Y}(\overline{y}) = 2$?

a) An ideal is prime if and only if the quotient ring it defines is a domain. In this case, the quotient is $\dfrac{k[x,y,z]}{(xy-z^2)}/(\overline y, \overline z)\cong k[x]$, which is clearly a domain.

b) $X\setminus Y$ is the open complement to $Y$ in the Zariski topology. Looking at $X, Y$ as classical varieties (Hartshorne says "set theoretically"), we see that $Y$ is defined by $\overline y=0$, which implies that the complement is $X\setminus Y = X\setminus V(\overline y) = D(\overline y)$. The topology of $X$ is determined set-theoretically, thus the complement of the scheme theoretic divisor of $\overline y$ is still determined by the same subset.

Thus, $X\setminus Y$ is defined by the localization $\left(\dfrac{k[x,y,z]}{(xy-z^2)}\right)_{\overline y} = \dfrac{k[x,y, y^{-1},z]}{(xy-z^2)}\cong k[y,y^{-1},z]$ at the function $\overline y$.

c) The divisor of $\overline y$ is given by the vanishing of $\overline y$, which we can view as the quotient $\dfrac{k[x,y,z]}{(xy-z^2)}/(\overline y) = \dfrac{k[x,z]}{(z^2)}\cong k[x]\otimes_k k[z]/(z^2)$. But this is just $Y$ with multiplicity two given by the nilpotent $z$ term.