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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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    And just to make sure that I understand: we are not talking about normal subgroups of an abelian group here, right?2012-05-21
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    @Thomas I'm not sure I understand the question... Isn't every subgroup of an abelian group normal? Anyway, the question is about rings.2012-05-21
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    Absolutely right. And you did write "ring". It most be a work injury that whenever I see a $\lhd$ I think normal subgroup... oops2012-05-21
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    @Thomas I think $\lhd$ is a pretty standard notation for "is an ideal of". Have you not encountered it? If it's not as common as I thought, I'll edit the question to make clear what I mean.2012-05-21
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    I actually have never seen it used for ideals before, but you don't have to edit your question. Looking at it again, it is clear what you are asking about.2012-05-21
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    @Thomas I have started a [meta question](http://meta.math.stackexchange.com/questions/4276/is-lhd-common-notation-for-is-an-ideal-of) to have this clarified. (I just thought you might be interested.)2012-05-21
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    A somewhat related question is [here](http://math.stackexchange.com/questions/144022/subideals-of-an-ideal/144027#144027). I can't resist echoing the factoid that idealhood is transitive in von Neumann regular rings.2012-05-21
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    @rschwieb I have already linked to this in the question.2012-05-21
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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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    Thank you for the answer. I don't require my rings to have a unity. (I think it would be a pretty unnatural assumption here too.)2012-05-21
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    What happens if the ring must have $1$. Is the statement also false?2015-04-21