If $R$ is a ring, J is an ideal in $R$, and $I$ is an ideal of $J$ (with $J$ considered as a ring), does it follow that $I$ is an ideal of $R$? That is, is $I$ necessarily closed under multiplication by elements of $R$? Surely $I$ is closed under multiplication by elements of $J$ (since $I$ is an ideal of $J$). The "obvious" approach to proof fails so quickly that it must be false, that "an ideal of an ideal is not necessarily an ideal of the big ring". Please provide a counterexample. (Or a proof?)
subideals of an ideal
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ring-theory
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1Fun fact: for a von Neumann regular ring $R$, being an ideal *is* transitive in the sense that $I\lhd J\lhd R$ implies $I\lhd R$. – 2012-05-11
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0It was very surprising for me to learn that such a result is true for closed ideals of C*-algebras. IIRC one uses approximate identities. – 2015-04-22