TITLE: subideals of an ideal
QUESTION [15 upvotes]: 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?)

REPLY [2 votes]: I can provide a commutative counterexample: Let $R=\mathbb Q[X]/(X^2)$, and $I=(\bar X)$. Then the ideals of $R$ contained by $I$ are simply $I$ and $0$. But any additive subgroup of $I$ is an ideal of $I$ (since the multiplicative structure on $I$ is the trivial one).