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Suppose $R$ is a commutative ring with 1. There are some statements that tells us if prime ideals behave in certain way, then all the ideals will behave in that way. For example,

If every prime ideal in $R$ is finitely generated, then every ideal in $R$ is finitely generated.

Similarly, we have

If every prime ideal in $R$ is principal, then every ideal in $R$ is principal.

I am interested in understanding what happens when we consider ideals that are 2-generated (i.e. they have two generators). So my question is:

If every prime ideal is 2-generated, then is it true that every ideal in $R$ needs less than or equal to 2 generators?

If the answer is affirmative, I guess the natural generalization has also answer "yes".

If every prime ideal is $k$-generated, then is it true that every ideal in $R$ needs less than or equal to $k$ generators?

Thanks!

Prism
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    If you like this idea then you have to read this paper by Lam and Reyes. Actually, I have a hard time not recommending this paper to everyone who knows what a prime ideal is... – rschwieb Aug 01 '13 at 16:57
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    @rschwieb: Thanks for this interesting suggestion. So they are giving far-reaching generalization of the first two statements in my post. I shall read it! – Prism Aug 01 '13 at 17:16
  • When you finish it, read this followup paper also :) – rschwieb Aug 01 '13 at 17:19
  • @rschwieb: Haha if I can finish it... That would be very first arxiv paper for me to read! Exciting! :) – Prism Aug 01 '13 at 17:23
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    $\mathbb{Z}[X]$ should be a counterexample. If I remember correctly, every prime ideal has one of the forms $(p)$, $(f)$ or $(p,f)$ where $p \in \mathbb{Z}$ is prime and $f$ is irreducible. In contrast to that, for any positive integer $n$ there is an ideal which needs $n$ generators. – Dune Aug 01 '13 at 19:06
  • @Dune: Could you elaborate on the last sentence? How do we construct an ideal which needs $n$ generators? – Prism Aug 01 '13 at 19:20
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    @Prism http://math.stackexchange.com/questions/110458/construct-ideals-in-mathbb-zx-with-a-given-least-number-of-generators –  Aug 01 '13 at 19:52
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    @Dune I think since you're the first to bring up this nice example, you should have the credit of writing it up as a solution. Please do this so we can get it out of the unanswered queue :) You might also incorporate a pointer to the fact about the prime ideals of $\Bbb Z[x]$. – rschwieb Aug 01 '13 at 20:01
  • @rschwieb,YACP: Thank you for the links! I mentioned them in my answer. – Dune Aug 01 '13 at 20:24

1 Answers1

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This answer is a summary of the above comments and some other threads mentioned there.

The ring $\mathbb{Z}[X]$ provides a counterexample to your question. To be precise, every prime ideal $\mathfrak{p} \subset \mathbb{Z}[X]$ has one of the forms: $(0)$, $(p)$, $(f)$ or $(p,f)$ where $p$ is a prime number and $f$ is an irreducible polynomial (even modulo $p$). A proof has already been given here. Particularly, every prime ideal is 2-generated.

On the other hand, the minimum number of generators of an arbitrary ideal is unbounded. In fact, for every positive integer $n$ there is an ideal $I \subset \mathbb{Z}[X]$ which needs exactly $n$ generators. Consider for example the ideal $I := (p,X)^n = (p^n, p^{n-1}X, \dots, pX^{n-1},X^n)$ which cannot be generated by fewer than $n+1$ elements. A proof has been given here.

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Dune
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