3

A commutative unital ring $R$ is called finitely cogenerated if for every desending chain of ideals $I_1\supset I_2\supset\cdots$ such that $\displaystyle\bigcap^\infty_{n=1}I_n=\{0\}$, we have $I_n=\{0\}$ for some $n\in\mathbb{N}^*$. An example of finitely cogenerated ring is constructed here.

I was wondering if an integral domain that is not a field can be finitely cogenerated. Of course, an integral domain that is not a field cannot be Artinian, which is stronger than being finitely cogenerated. Clearly, the example in the above link is not an integral domain because it has many zero divisors. I suppose that the answer is negative, since we must have $\displaystyle\bigcap^\infty_{n=1}I^n\neq\{0\}$ for every nonzero ideal, because an integral domain has no nonzero nilpotent ideal. This is a much too strong condition. Thank you for your help in advance.

Jianing Song
  • 2,545
  • 5
  • 25

1 Answers1

1

The definition of "finitely cogenerated" that I am acquainted with does not require the indexing set to be countable. So to me, a (right) finitely cogenerated ring is one in which satisfies either of the two equivalent conditions:

  1. For every indexed collection of right ideals such that $\cap T_{i\in I }=\{0\}$, there is a finite subset $F$ of $I$ such that $\cap T_{i\in F }=\{0\}$
  2. For every indexed collection of right ideals forming a chain such that $\cap T_{i\in I }=\{0\}$, there is an index $j\in I$ such that $T_j=\{0\}$

By this definition, a finitely cogenerated integral domain would have to be a field: In an integral domain, the collection of all nonzero ideals either

  1. has trivial intersection, in which case it's obvious there is no finite subcollection intersecting to $\{0\}$, and it's not f.cog; or
  2. has nontrivial intersection, in which case that intersection is a minimal ideal, in which case the ring is a field.

It isn't clear to me that the definition you gave is equivalent. I think there are valuation domains whose ideals (excluding $\{0\}$) are order isomorphic to totally ordered sets in which you can't find a countable cofinal subset, and those would be vacuously f.cog in your definition.

rschwieb
  • 160,592
  • 1
    Ah yes, I missed that the definition does not require the family to be countable, and your comment on valuation domains is an interesting idea! Thanks! – Jianing Song Oct 25 '24 at 20:31
  • @JianingSong You might find some more interesting examples to look at on my site. – rschwieb Oct 25 '24 at 20:57
  • Thanks! That's like a topological $pi$-base for ring-theory :) – Jianing Song Oct 25 '24 at 21:08
  • @JianingSong Yes... I've corresponded with the creators of π-base since it before when it used to be called "spacebook," IIRC – rschwieb Oct 25 '24 at 21:12
  • Great work! By the way, I have found indeed an example showing that the arbitrary family cannot be replaced by a countable family in the definition :https://math.stackexchange.com/q/1457091. – Jianing Song Oct 25 '24 at 22:17
  • And an example in the equivalence of finitely generated modules: https://mathoverflow.net/q/88608 (always the same idea). – Jianing Song Oct 25 '24 at 22:27
  • 1
    @JianingSong Yep, as I suspected. Good job finding both those questions! – rschwieb Oct 26 '24 at 16:11