First, some setup:

So: given a commutative ring $R$, let $Ideals(R)$ be set of ideals of $R$ and let $IdealClosure(R)$ be the set of closure operations $cl: \mathcal{P}(R) \rightarrow Ideals(R)$. In other words, let $IdealClosure(R)$ be the set of functions $cl: \mathcal{P}(R) \rightarrow \mathcal{P}(R)$ such that

  • $cl(A) \subseteq R$ is an ideal for all $A \subseteq R$,
  • $A \subset cl(A)$ for all $A \subseteq R$,
  • $cl(cl(A)) = cl(A)$ for all $A \subseteq R$, and
  • $A_1 \subseteq A_2 \implies cl(A_1) \subseteq cl(A_2)$ for all $A_1,A_2 \subseteq R$.

Note that since the ideals of a ring form a bounded lattice, the set $IdealClosure(R)$ is also a bounded lattice. Now if $\mathcal{R}$ is some set of rings, let $IdealClosure(\mathcal{R})$ be the collection of functions which assign to each ring $R$ an element of $IdealClosure(R)$. Again, $IdealClosure(\mathcal{R})$ is a bounded lattice by defining the lattice operations pointwise, ring by ring.

Let $\mathcal{R}$ be a set containing at least one representative from each isomorphism class of the finitely presented commutative rings. Then we can assume $\mathcal{R}$ is countable.

It seems then that given a few more 'naturality' conditions the set $IdealClosure(\mathcal{R})$ would likewise countable. For example, without going to far into symbolic logic, a natural ideal closure operator would be defined by one of a countable collection of statements in FOL or ZFC.

One obvious condition is this: an ideal closure operation should treat isomorphic rings equivalently, or more formally: if $R_1 \in \mathcal{R}$, $R_2 \in \mathcal{R}$, $\varphi: R_1 \rightarrow R_2$ is an isomorphism, and $cl_1 = c(R_1)$, $cl_2 = c(R_2)$ where $c \in IdealClosure(\mathcal{R})$ then $\varphi[cl_1(A)] = cl_2[\varphi(A)]$ for all $A \subseteq R$.

What other conditions are reasonable? Are such sets of closure operations studied?


1 Answer 1


Note that for any such "closure operations" one has $cl(A) = cl(\langle A\rangle)$ where $\langle A\rangle$ denotes the ideal generated by $A$. Hence it can be defined as an operation on ideals.

Now if you add a conditions of compatibility with intersection of ideals: $cl(A \cap B) = cl(A) \cap cl(B)$ for $A$ and $B$ ideals then a closure operation on a ring $R$ is the same as a nuclei on the set of ideals of $R$, i.e. it is the same as a sub-locale of the spectrum of the ring (i.e. a subspace of the spectrum from the point of view of pointfree topology).

With this insight in mind it is fairly easy to see that this has no chance to be countable: Pick any subset of prime numbers $Q$, then there is a sub-locale $Q \cup \{ \infty\}$ of $Spec \mathbb{Z}$ which will corresponds to such a closure operation on $\mathbb{Z}$:

To a set $P \subset \mathbb{Z}$ you associate the intersection of $\mathbb{Z}$ with the ideal generated by $P$ in $\mathbb{Z}[Q' ^{-1}]$ where $Q'$ is the complement of $Q$.

Now, for any ring $A$, $Spec A$ has a canonical map to $Spec \mathbb{Z}$, so you will obtain a sublocale of $Spec A$ by just pulling back any given sublocale of $Spec \mathbb{Z}$ hence any sublocale of $Spec \mathbb{Z}$ gives a canonical and functorial choices of such a closure operation on $Spec A$.

For example, here you will obtain on every ring $A$ the closure operation which associate to any set $P \subset A$ the intersection of $A$ with the ideal of $A[Q'^{-1}]$ generated by $P$.

Of course the set of subset of of prime number is uncoutable so you will obtain an uncountable number of such assignment of closure operations.

Note that if you take both the "intersection" condition and sufficiently* strong functoriality condition on your assignment in the definition you can have that a functorial assignment of closure operation is the same as a closure operation on the base ring, and hence if you chose a base ring with a finite spectrum then you can have only a finite set of such functorial assignment of closure operation.

(*) meaning that when you have a ring morphism $f :A \rightarrow B$ then the closure operation on $B$ is given by the closure operation on $A$ in the way I described above for $\mathbb{Z}$ and an aribtrary ring $A$.

  • $\begingroup$ That is a good argument to demostrate that there may be uncountably many closure operations. I meant to suggest only that with more conditions the set would be countable. If we are doing commutative algebra in ZFC then certainly there are only countably closure operations one can even define. I would find it fascinating if one could give some conditions, native to commutative algebra (and w.o. any high powered logic notions), that would whittle down the number of options down to a countable number. $\endgroup$
    – Rex Butler
    Jul 22, 2015 at 20:08
  • $\begingroup$ Then I would say your question is too vague: there is plenty of way to do such a thing and we don't know what you want of such a notion... You can consider closure operation which are "finetely generated" in the sense that there is a finite set $P$ of element of various rings such that taking the closure of a set is to take the ideal generated by the set plus the image of element of $P$ along morphism/isomorphism. Or you can define a notion of "computable/recursive" closure operation based on the notion of recursive function etc... $\endgroup$ Jul 23, 2015 at 8:36
  • $\begingroup$ Clearly I need to learn more commutative algebra before I proceed with such a question. Thanks for your response. $\endgroup$
    – Rex Butler
    Jul 23, 2015 at 15:43
  • 1
    $\begingroup$ If you want to get a locale you must only consider radical ideals. $\endgroup$ Dec 13, 2021 at 21:16

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