I am just getting started on spectrums of rings. I see how it is natural to augment the set of prime ideals with the Zariski topology, but from my poor intuition on the topic I don't see how any of the points in such a topology would fail to be closed, or limit points. Is there any ring whose spectrum contains points that are not closed? Or is this part of the point of the definition to only retain closed points in the topology?
2026-03-25 12:26:10.1774441570
Is every point in the spectrum of a ring $R$ closed?
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Suppose that $R$ does not have divisor of $0$, thus the ideal $(0)$ is prime and its closure is $Spec(R)$. $(0)\subset V(I)$ implies that $I\subset (0)$ and $I=0$, if $Spec(R)$ has more than two point, $(0)$ is not closed. For such a $R$ (without divisors of zero), $(0)$ is called the generic point of $Spec(R)$.