1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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 $S=V(I)$. $\endgroup$ – Mindlack Nov 27 '19 at 13:31
2
$\begingroup$

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.

To see that closed points give maximal ideals, just notice that all ideals are contained in maximal ideals. Therefore, all closed sets contain at least one maximal ideal. So if a single point is a closed set, it has to be a maximal ideal.

$\endgroup$
1
$\begingroup$

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.

Here's a way to prove that closed points correspond to maximal ideal. Actually, one has the following result:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.