A commutative unital ring $R$ is called finitely cogenerated if an intersection of ideals of $R$ is zero, then a finite intersection of them is also zero. I am looking for a finitely cogenerated commutative unital ring $R$ with infinitely many maximal ideals.


1 Answer 1


Take $R=\mathbb Z\oplus \mathbb Z_{p^{\infty}}$. This means: (1) Start with the Prufer group $\mathbb Z_{p^{\infty}}$ for some prime $p$ and make it a commutative nonunital ring by equipping it with zero multiplication. (2) Formally adjoin a unit element by taking the direct sum $\mathbb Z\oplus \mathbb Z_{p^{\infty}}$ of additive groups and defining multiplication by $(a,u)\cdot(b,v) = (ab,av+bu)$.

Certainly $R$ is a commutative ring with unit $(1,0)$. Also, the function $(a,u)\mapsto a$ is a homomorphism from $R$ onto $\mathbb Z$. Since the latter has infinitely many maximal ideals, the former must also have infinitely many. To finish showing that $R$ satisfies all the conditions it suffices to explain why the following is true:

Claim. If $\{I_k\;|\;k\in K\}$ is a set of ideals and $\bigcap_{k\in K} I_k$ is zero, then one of the ideals $I_k$ is already zero.

Proof of Claim. (Contrapositive)

Note that $N = \{0\}\oplus \mathbb Z_{p^{\infty}}$ is an ideal which squares to zero, and its subideals are exactly those of the form $\{0\}\oplus H$ where $H$ is an additive subgroup of $\mathbb Z_{p^{\infty}}$.

Let me argue that if $I$ is a nonzero ideal of $R$, then $I\cap N$ is also nonzero, hence $I\cap N = \{0\}\oplus H$ for some nonzero subgroup $H\leq \mathbb Z_{p^{\infty}}$. To this end, suppose that $(a,u)\in I-\{(0,0)\}$. If $a=0$, then $(0,0)\neq (a,u)=(0,u)\in I\cap N$, and we are done in this case. In the alternative case where $a\neq 0$, there is some $v\in\mathbb Z_{p^{\infty}}$ such that $av\neq 0$, hence $(0,0)\neq (0,av)=(a,u)\cdot (0,v)\in I\cap N$. Our assertion is established.

Now, $\mathbb Z_{p^{\infty}}$ is a subdirectly irreducible abelian group, so it has a least nonzero subgroup $H$, and by the result of the previous paragraph $\{0\}\oplus H$ is contained in every nonzero ideal of $R$. Hence if all $I_k, k\in K$, are nonzero, then $\bigcap_{k\in K} I_k\neq 0$. \\\

  • $\begingroup$ $R$ is the idealization of the $\mathbb Z$-module $\mathbb Z_{p^{\infty}}$, and its (maximal) ideals are easily determined. $\endgroup$
    – user26857
    Sep 27, 2016 at 19:44

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