To sum up the comments by Keenan and Georges, generic points are minimal prime ideals. It seems that you are asking for a closed set $V(I)$ such that every prime ideal in $V(I)$ is minimal.
Since no two points in $V(I)$ are comparable in $\subseteq$ (otherwise we contradict minimality) we have to have that every point is maximal. Since the intersection of two prime ideals containing $I$ is also a prime ideal containing $I$ then we have to have that $V(I)$ is a singleton to begin with.
Assuming, however, that you confused the double-negations a bit let me answer a slightly different question. Without the axiom of choice there exists a [commutative unital] ring $R$ and a decreasing chain of prime ideals in $R$ without a minimal element.
To see a wonderfully vivid proof that the existence of a minimal prime ideal implies the axiom of choice see this MathOverflow answer.
Therefore assuming choice fails there is such a ring that has a decreasing chain of prime ideals without a minimal element, as promised.
To the edit: much like the part above, the existence of maximal ideals in commutative unital rings is also equivalent to the axiom of choice (see this). Namely without the axiom of choice there is a ring with a chain of ideals without a maximal element.
I am unsure, however if you can find a "maximal" prime which is not maximal and not contained in any real maximal ideal.