For a commutative unital ring $R$, let $J(R)=0$ (a semiprimitive ring), and for any family of finitely generated ideals $\{I_i\}$ if $\cap I_i=0$, then a finite intersection of $\{I_i\}$'s is also zero. Is there any characterization for such a ring?
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4$\begingroup$ An $R$-module $M$ is called finitely cogenerated if for every family of submodules $(M_i)_{i \in I}$ with $\bigcap_{i\in I} M_i=0$ there is some finite subset $J \subseteq I$ such that $\bigcap_{i \in J} M_i=0$. In your case, $M=R$. However, the property you mention is slightly weaker since you only allow finitely generated $M_i$. $\endgroup$– HeinrichDCommented Sep 26, 2016 at 19:00
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1$\begingroup$ This condition is weaker than being finitely cogenerated: for instance, let $K$ be a field, $V$ an infinite-dimensional vector space, and $K_V$ the algebra $K\oplus V$ with multiplication $(t,v)(t',v')=(tt',tv'+t'v)$. Then its ideals $\neq K_V$ are the subspaces of $V$; the finitely generated ideals are the finite-dimensional subspaces. The second, but not the first, satisfy the intersection property. $\endgroup$– YCorCommented Sep 27, 2016 at 21:50
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1$\begingroup$ Nevertheless, the weak version of being finitely cogenerated implies the property (in an arbitrary module over a unital associative ring) P that every nonzero submodule contains a simple (=minimal) submodule. Given a module $M$ with socle $S(M)$ (the submodule generated by simple submodules), this amounts to the condition that $S(M)$ intersects nontrivially every nonzero submodule. Then the condition is that the socle itself has the weak finite cogeneration property. $\endgroup$– YCorCommented Sep 27, 2016 at 21:53
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