# Is the Zariski topology on the spectrum of a ring defined for a fixed ideal, or all ideals?

Initially this question was meant to ask how to prove that closed points of $$\text{Spec}(R)$$ correspond to maximal ideals of $$R$$ (This question had been asked previously in various forms, e.g. here and here, but I hadn't found one that explained it satisfyingly without relying on the theory of schemes, or anything more than elementary ring theory).

My initial understanding was that the Zariski topology on $$\text{Spec}(R)$$ is that closed sets are sets of prime ideals which contain an ideal $$I$$, for all ideals $$I$$ of $$R$$. However, after considering how to prove the above claim (specifically that closed points give maximal ideals), I think this definition is supposed to be that for some fixed ideal $$I$$, the topology is defined. Wikipedia, the texts I have read from, and other resources, somehow do not make the distinction (or maybe I lack the comprehension skills).

So, is the latter definition the correct one?

Specifically, in the other direction, I am able to accept the conclusion only if the latter definition is correct. If $$\mathfrak{p} \in \text{Spec}(R)$$ is closed, then it is equal to its closure, i.e. the smallest closed set containing $$\mathfrak{p}$$, and by the definition of the closed sets, $$\mathfrak{p}$$ is maximal.

If I use the first definition, it does not seem possible to prove the claim, unless I am missing something incredibly easy.

• The definition is as follows: a subset $S$ of the spectrum of $A$ is Zariski-closed if there is an ideal $I$ of $A$ such that $S=V(I)$. – Mindlack Nov 27 '19 at 13:31

Yes, a closed set in the Zariski topology is a set of the form $$V(I) = \{\mathfrak{p}\in\rm{Spec}(R) \mid I\subset \mathfrak{p}\}$$. If you fix the ideal $$I$$, then there is only one closed set in the topology! That clearly isn't right.
Even for a fixed closed set in $$\DeclareMathOperator{\spec}{Spec}\spec R$$, the defining ideal is not unique, since $$V(I)=V(\sqrt I)$$ by Hilbert's Nullstellensatz.
The closure of a point $$\mathfrak p$$ in $$\spec R$$ is $$\enspace\overline{ \{\mathfrak{p}\}}=V(\mathfrak p)$$.
Indeed, if $$\mathfrak q\in \overline{ \{\mathfrak{p}\}}$$, this means any elementary open set $$D(f)$$ which contain $$\mathfrak q$$ contains $$p$$, i.e. if $$f\notin\mathfrak q\implies f\notin\mathfrak p$$, in other words $$R\smallsetminus\mathfrak q\subseteq R\smallsetminus\mathfrak p\iff \mathfrak q\supseteq\mathfrak p.$$ Now if $$\mathfrak p$$ is closed, by the previous result, $$\:V(\mathfrak p)=\{\mathfrak p\}$$, which means $$\mathfrak p$$ is a maximal ideal.