Let $R$ be a commutative ring and $ \operatorname{Spec}(R)$ is the set of prime ideals with Zariski toplogy.
The set of maximal ideals is denoted by $X= \operatorname{mSpec}(R)$.
We consider $\overline{X}$, which is the Zariski closure of $X$.
Can you describe necessary and sufficient conditions for $X$ to satisfy $\overline{X}=X$ ?
2026-03-25 11:03:22.1774436602
Zariski closure of the set of maximal ideals
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If $R$ is Noetherian, then $X$ is closed iff $R$ has only finitely many maximal ideals. The reverse implication is trivial; for the forward implication, note that if $X$ is closed then it is Spec of a quotient of $R$ which is zero-dimensional and Noetherian, but a zero-dimensional Noetherian ring is Artinian and has only finitely many maximal ideals.