In $\mathbb{Z}$, or more generally a Dedekind domain $D$, an element $x$ is uniquely determined by its image in the associated prime fields $D/\mathfrak{p}D$ as $\mathfrak{p}$ varies over all prime ideals.
How generally does this hold? Do we require that for a commutative ring $R$ to have this property, it be a UFD?
We may alternatively think of an element $x\in R$ as a section of the structure sheaf of Spec($R$). The question is then whether this identification is 1-to-1
This works in general for reduced rings (the nilradical is $(0)$). To see this, if you have $x,y \in D$ such that their image in every quotient by a prime ideal is the same, then $(x-y)$ is in every prime ideal and hence in the nilradical, so $x = y$.
The same idea works with maximal ideals if your ring is Jacobson and reduced (or more weakly, the Jacobson radical is $(0)$). I mention this to avoid silly examples with domains, since $(0)$ is a prime ideal and then the result follows trivially.
You can also prove that the above are necessary and sufficient. If you have a non-zero nilpotent element $e$, you can consider $x + e$ for some $e$ and get the same image in all prime ideals. Similar to prove that the Jacobson radical needs to be zero if an element is to be determined by the image in the quotient fields.