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$\begingroup$

$\lhd$ will stand for "is an ideal of" in this post.

Let $R$ be a commutative ring, $J\lhd I\lhd R$. Does it follow that $J\lhd R?$

I don't think it does, but I'm having difficulty finding a counterexample. An example for a non-commutative $R$ is here.

I've tried several things, but at random really, so I don't think it makes sense to post it here.

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    Doh I overlooked that somehow!2012-05-21

1 Answers 1

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Consider everyone's favourite commutative ring $R=\mathbb{Z}[x]$. Let $I=\langle x^2\rangle\triangleleft R$ and $J\subseteq I$ be the subset of those polynomials that don't contain a $x^3$ term. Clearly $J$ is an ideal of $I$, since you can't produce a $x^3$ term from terms of degree greater than 2, but $J$ isn't an ideal of $R$.

Added: of course, this example works if you don't require your rings to have a unit. I'd have to think some more in the other case.

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    What happens if the ring must have $1$. Is the statement also false?2015-04-21