5
$\begingroup$

let $K$ be a number field of degree $d$ over $\mathbb{Q}$), Let $\mathcal{O}\subset K $ be an order (i.e. a $\mathbb{Z}$-lattice of $K$ contained in the integer ring $\mathcal{O}_K$ of $K$). If $ \mathcal{O}= \mathcal{O}_K $, we know that any maximal ideal in $ \mathcal{O}_K $ can generated by 2 element. So what's the case in general?

Maybe I should ask a more general question, let $A$ be $\mathbb{Z}$-algebra, and finite free of rank$=d$ as a $\mathbb{Z}$-module, what can we say about the number of generators of maximal ideals of $A$ (in terms of $d$), for example, we have a trivial upper bound $d.$

PS: I asked this question on MSE,https://math.stackexchange.com/questions/1486422/number-of-generators-of-maximal-ideals-in-an-order-of-a-number-field but got no answer.

A special case of the general question, say $A$ is an order in a finite product of field $\prod_{i=1}^n K_i$, i.e. $A\subset \prod \mathcal{O}_{K_i} $ with rank $d=\Sigma [K_i:\mathbb{Q}]$, what's the upper bound of number of generators of maximal ideals of $A$ (in terms of $d$ and $n$). If $A$ is the product of orders in all fields, then we can say something easily (see the answer by @Matthias Wendt). So maybe the question is the structure of $A$?

Improve this question
$\endgroup$
4
  • $\begingroup$ What do you mean by a $\mathbb Z$-algebra? Usually algebras are defined over fields. $\endgroup$
    – Kimball
    Oct 21, 2015 at 16:08
  • $\begingroup$ @Kimball: The definition I'm familar with is: An extension $S$ of $R$ of rings $R,S$ is a ring homomorphism $R \to S$. If $R$ is in addition commutative, $S$ is called an $R$-algebra. $\endgroup$
    – Todd Leason
    Oct 21, 2015 at 20:29
  • 1
    $\begingroup$ In the $R$-algebra $A=R[x_1,x_2,...,x_n]/J^2$ where $J=(x_1,x_2,..,x_n)$ the image of $J$ in $A$ needs $n$ generators, and $A$ is free over $R$ of rank $n+1$, so basically the trivial upper bound is essentially the best you can do in the generality you've written your question. $\endgroup$
    – eric
    Oct 21, 2015 at 20:41
  • $\begingroup$ Please edit in a link to the m.se post, and also put a link to this post there. $\endgroup$
    – Gerry Myerson
    Oct 21, 2015 at 22:27

1 Answer 1

Reset to default
4
$\begingroup$

I don't know of any general statement. A bit of digging turned up the following paper:

  • C. Greither. On the two generator problem for the ideals of a one-dimensional ring. J. Pure Applied Alg. 24 (1982), 265-276.

Theorem 3.6 in there states that an order $R$ in a number field $K$ has the property that every ideal is generated by $2$ elements if the discriminant does not contain a fourth power. Example 3.8 gives $\mathbb{Z}[2\cdot\sqrt[3]{5}]$ as an example that does not have this property. The paper also contains statements on group rings, and of course links to further literature; apparently, the question of 2-generator rings was well-studied at some point.

More general results would most likely depend on the structure of the singularities of the order. For example, the function ring of an affine curve has the 2-generator property if and only if the singularities are double points. This follows from Theorem 2.1 in the following paper (probably already from the referenced work of Bass...):

  • L.S. Levy, R. Wiegand. Dedekind-like behaviour of rings with 2-generated ideals. J. Pure Applied Alg. 37 (1985), 41-58.
Improve this answer
$\endgroup$
1
  • $\begingroup$ But indeed “ the discriminant does not contain a fourth power” is too strong! $\endgroup$
    – user42690
    Oct 21, 2015 at 21:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.