The example is about computing the Weil divisor class group of the affine cone Spec$k[x,y,z]/ \langle xy-z^2\rangle$.
I can show that the group is cyclic generated by the divisor $V(y,z )$ and that this is principle in order $2$, so it remains to show that the divisor $V(y,z)$ is not itself principal.
Hartshorne says
since $A$ is integrally closed, it is equivalent to show that the prime ideal of $Y$, namely $\mathfrak{p} = \langle y,z \rangle $ is not principal.
My problem is that I don't see why being integrally closed makes these two things equivalent.
Suppose $A$ is an integrally closed domain and let $Z$ be a principal prime Weil divisor with generic point $\mathfrak{p}$. Then there is an element $f \in K(A)$ such that $f$ generates the maximal ideal $\mathfrak{p}A_{\mathfrak{p}}$ in $A_{\mathfrak{p}}$. Moreover, $f$ lies in the other local rings $A_{\mathfrak{q}}$ whenever $\mathfrak{q}$ has height $1$. So since $A$ is integrally closed, this gives us that $f \in A$. But I don't see how this is equivalent to showing that $f$ generates $\mathfrak{p}$.
I try to show this: Certainly $\frac{f}{1}$ generates $\mathfrak{p}A_{\mathfrak{p}}$ by assumption. So let $x \in \mathfrak{p}$. We want to be able to write $x$ as a multiple of $f$. In the local ring, $\frac{f}{1}\cdot \frac{a}{s} = \frac{x}{1}$. Since $A$ is a domain, this gives us that $fa = sx$, which is not sufficient to show that $f$ generates $\mathfrak{p}$.
Is someone able to point me in the right direction?
Two crucial facts you haven't used yet: $v_p(f) = 1$ and $v_q(f) = 0$ for all height 1 prime ideals $q \neq p$.
Now, let $q$ be a height one prime prime ideal, and we have $v_q(f) + v_q(a) = v_q(s) + v_q(x)$. Since $v_q(x) \geq 0$ and $v_q(f) = 0$, we have $v_q(a) \geq v_q(s)$
For $p$, we have $1 + v_p(a) = v_p(x)$. Since $v_p(s) = 0$ and $v_p(x) \geq 1$, $v_p(a) \geq 0 = v_p(s)$
Therefore, $div(a/s) \geq 0$ so $a/s \in A$. Your equation is now $f * \frac{a}{s} = x$ which takes place in $A$ (here we use that $A$ is an integral domain, so $A \rightarrow A_p$ is injective)
See the proof of proposition 6.2 and adapt it to this case.