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In an algebraic number theory course, my lecturer said that any ideal of $\mathcal O_K$, where $K$ is a quadratic number field, is generated by at most two elements. I am wondering why this is. When $K$ is as above, $\mathcal O_K$ is a free $\mathbb Z$-module of rank 2, so I want to say that if an ideal was generated by more than two elements and couldn't be "reduced" to a form that is generated by two elements, then it would be a free $\mathbb Z$-module of rank $>2$, a contradiction.

I am wondering if this logic is justified, in particular:

Is the minimal number of generators of an ideal of $\mathcal O_K$ equal to the rank of the ideal as a free $\mathbb Z$-module?

(If the result is true, this would also give a solution to a generalised result about non-quadratic number fields $K$.)

I can't think of how to start with this, as I have trouble thinking about non-principal ideals.

(My lecturer also mentioned that for any $K$, an ideal of $\mathcal O_K$ will still be minimally generated by at most two generators, but that this is a deep theorem. If you can give me a reference to where this is proved I would be very grateful, though this is not my main question.)

James
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  • $O_K$ is a Dedekind domain, and the ideals of Dedekind domains are $2$-generated. – user26857 May 17 '15 at 21:54
  • Why is this? This may be the result my lecturer was referring to (see final paragraph) – James May 17 '15 at 21:55
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    See http://math.stackexchange.com/questions/597543/in-a-dedekind-domain-every-ideal-is-either-principal-or-generated-by-two-element – SashaP May 17 '15 at 21:56
  • Your question is about the ideals of $O_K$ or in general? – user26857 May 17 '15 at 21:58
  • The question is about $\mathcal O_K$ but if it can be answered more generally I'm happy to deal with a more general case – James May 17 '15 at 21:59
  • It's not clear to me what kind of $O_K$ you are using, but it seems that these are of the form $\mathbb Z[\sqrt d]$ or so. (You mentioned rank 2, that's why I think so.) In this case an ideal can't have exactly two generators (not less), and rank one as a free module since then it must be principal. – user26857 May 17 '15 at 23:04
  • @user26857 $K$ is a quadratic number field so $\mathcal O_K$ is $\mathbb Z[\sqrt d]$ or $\mathbb Z\left[\frac{1+\sqrt d}{2}\right]$. Why can't an ideal have 2 generators? e.g. $(2, \sqrt {-7})$ works, no? – James May 18 '15 at 00:06
  • I didn't say it can't have two generators! What I wanted to say is that if $I$ has two generators then its rank as a free $\mathbb Z$-module can't be one. – user26857 May 18 '15 at 00:22
  • @James I deleted my answer because it doesn't really answer the question in yellow, it just explains the reasoning behind the argument. Let me know if you want me to undelete it. Doesn't matter to me one way or another. – Matt Samuel May 18 '15 at 00:38
  • @MattSamuel I was sleeping so I didn't get a chance to see your answer. Anything might be helpful to me, so you may as well undelete it in case it is. If you want you could put some kind of disclaimer at the start saying "this doesn't answer the main question but..." – James May 18 '15 at 09:10
  • The unit ideal is generated by 1 element. But the unit ideal is the whole ring, which in general will not have rank 1 over $\mathbb{Z}$. – User0112358 Aug 22 '18 at 01:41

3 Answers3

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In response to your reference questions, Daniel Marcus provides a nice proof (see his text 'Number Fields', Theorem 17) that any ideal in a number ring (thus the number ring itself) can be minimally generated by at most two elements.

Arbutus
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An ideal is a $\mathbb{Z}$-submodule. It is possible for two elements in $\mathbb{Z}^2$ to generate a proper submodule, but it is not possible for three elements to be linearly independent over $\mathbb{Z}$ (we can always find a nontrivial linear combination that is $0$). Furthermore, any submodule is free and therefore has a linearly independent generating set. Thus an ideal has a minimal generating set with at most two elements because in fact it is additively generated over $\mathbb{Z}$ by a set with at most two elements.

This does not rule out a priori that there exists a minimal generating set with more than two elements, in the sense that none of them could be omitted and have the set still generate the ideal. That would just be the "wrong" generating set, and we could pick a smaller, completely different one.

Note this doesn't answer the main question.

Matt Samuel
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If I understood well your question, the answer is negative: in $\mathbb Z[i]$ the principal ideal $I=(2+i)$ is a free $\mathbb Z$-module of rank two.

On the other side, a two-generated ideal of a quadratic number field is necessarily a free $\mathbb Z$-module of rank two (why?).

user26857
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