Given $A$ a Dedekind ring, with $X=\operatorname{Spec}(A)$.
I want to understant why $\operatorname{Cl}(X)=\operatorname{Cl}(A)$ (Hartshorne page 132 example 6.3.1).
For that I need to verify that for all $\mathfrak{p}$ prime ideal of $A$, $\mathfrak{p}$ is principal iff $V(\mathfrak{p})=Y$ (the closed subscheme of $X$ associated to $\mathfrak{p}$) is principal.
It is clear to me that if $\mathfrak{p}$ is principal with $\mathfrak{p}=fA$ then $V(\mathfrak{p})$ is principal with $V(\mathfrak{p})=(f)$.
I don't see why the converse is true: taking $f\in A$ with $V(\mathfrak{p})=(f)$ then (because $v_Y(f)=1$) we have $f\in\mathfrak{p}A_\mathfrak{p}$ and $f\notin(\mathfrak{p}A_\mathfrak{p})^2$. As $A_\mathfrak{p}$ is regular of dimension 1 (DVR) one has $\mathfrak{p}A_\mathfrak{p}=fA_\mathfrak{p}$. I can deduce that $f\in\mathfrak{p}$ but not that $fA=\mathfrak{p}$. I don't used that for the other prime $Y'=V(\mathfrak{q})$ one has $V_{Y'}(f)=0$ that if $f\notin\mathfrak{q}$, but how? I feel I'm missing something obvious...