Let us say a given commutative ring $R$ has property $\mathcal{P}$ if for every two elements $f,g \in R\setminus 0$, there exists a prime ideal $q \subset R$ such that both $f \notin q$ and $g \notin q$.
Note: obviously if $R$ has nontrivial nilradical then $R$ doesn't have property $\mathcal{P}$.
I was wondering, can anyone provide an example of $R$ reduced (i.e. having trivial nilradical) but still failing to have property $\mathcal{P}$?
Also, does the above property hold if $R$ is a reduced algebra (unital, assoc, comm) over a field?
Note: my motivation for considering this property is, if I'm not mistaken: if $R$ has property $\mathcal{P}$ then the intersection of any two nonempty distinguished opens is nonempty. Since the distinguished opens form a base, it follows that every open set of $\mathrm{Spec}R$ is dense.
Let $F$ be your favorite field, and consider the $F$-algebra $R=F[x,y]/\langle xy\rangle$. This ring is reduced, but it does not have property $\mathcal{P}$, since every prime ideal of $R$ has to contain one of the non-zero elements $\bar{x}$ and $\bar{y}$.