9

$\lhd$ 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.

Community
  • 1
  • 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. –  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. –  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
  • @Thomas I have started a meta question to have this clarified. (I just thought you might be interested.) –  May 21 '12 at 22:57
  • A somewhat related question is here. I can't resist echoing the factoid that idealhood is transitive in von Neumann regular rings. – rschwieb May 21 '12 at 23:04
  • @rschwieb I have already linked to this in the question. –  May 21 '12 at 23:07
  • Doh I overlooked that somehow! – rschwieb May 21 '12 at 23:09

1 Answers1

7

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.

Miha Habič
  • 7,164
  • 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.) –  May 21 '12 at 21:42
  • What happens if the ring must have $1$. Is the statement also false? – user10024395 Apr 21 '15 at 11:29