Every irreducible affine scheme $\mathrm{Spec}(R)$ contains a generic point, namely $\eta:=\mathrm{Nil}(R)$. If $R$ is a domain then $\eta=(0)$. This is a point which is Zariski dense in $\mathrm{Spec}(R)$, meaning that $\eta$ is contained in every Zariski open subset of $\mathrm{Spec}(R)$.
I'm looking for a fairly elementary statement as follows.
Theorem. Let $R$ be a commutative domain. Then every prime ideal $p$ is X.
The rough proof sketch would be as follows:
- "X" is true for $p=0$.
- The set $V$ of "X" primes is Zariski closed.
- Since $0$ is dense, it must be that $V = \mathrm{Spec}(R)$, so every prime is "X".
The set $V$ is the titular "closed" condition.
It might be asking too much, but does anyone know such a statement whose proof follows the above sketch?
If the statement is sufficiently nontrivial, I imagine that Step 2 would be where most of the element-wise work is baked in (because you essentially have to show that $U=V^c$ is a union of "basic" open sets $X_f=\mathrm{Spec}(R_f)$, which is a localization type of thing). Or, it is already hard enough to show the result for $p=0$, and it is trivial to go from $p$ to $0$ by taking quotients.
Although the statement might end up being basically trivial, this "topological" perspective on it may be illuminating.