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$\lhd$ 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? –  Thomas May 21 '12 at 21:55
    
@Thomas I'm not sure I understand the question... Isn't every subgroup of an abelian group normal? Anyway, the question is about rings. –  user23211 May 21 '12 at 21:58
    
Absolutely right. And you did write "ring". It most be a work injury that whenever I see a $\lhd$ I think normal subgroup... oops –  Thomas May 21 '12 at 22:01
    
@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. –  user23211 May 21 '12 at 22:05
    
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. –  Thomas May 21 '12 at 22:10
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1 Answer

up vote 4 down vote accepted

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.) –  user23211 May 21 '12 at 21:42
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