Matrices $A$ and $B$ are integrally equivalent if there is an invertible integer matrix $L$ and $L^{-1}AL=B$. Suppose $f(t)$ is an integer polynomial with no repeated factors. Latimer and MacDuffee proved that the number of integral similarity classes of matrices with characteristic polynomial equal to $f(t)$ is equal to the number of non-singular ideal classes of $\mathbb{Z}[\theta]$. (And it's not clear to me what a non-singular ideal is, or was in 1933.)

If for each irreducible factor of $f$ the corresponding number field has class number 1, and if the ring of algebraic integers in it is equal to $\mathbb{Z}[\theta]$, then it follows that two matrices with characteristic polynomial $f$ are integrally equivalent. (This provides another way to verify Tracy Hall's computation as reported in example, which is what started me down this rabbit hole.) But the above two assumptions on $K$ are strong, and I am trying to find out what is known when these conditions are weakened. Hence:


  1. Can someone point me to a reference (or more) concerning ideal classes in $\mathbb{Z}[\theta]$ when this order is not maximal? Even just a proof of the fact that the number of ideal classes in $\mathbb{Z}[\theta]$ is finite? [I am assured that this is a fact, and it appears to follow from the usual proof for Dedekind domains; I am hoping that any source that treats this explicitly will offer further information.]

  2. Is there any characterization of the non-invertible ideals in $\mathbb{Z}[\theta]$? [I am aware of results in Harvey Cohn's "A Classical Invitation..." about ideals coprime to the conductor.]

  3. Will the theory simplify if I assume that $\theta$ is totally real?


I am interested in cospectral graphs - non-isomorphic graphs whose adjacency matrices are similar. Experimental evidence suggests that almost all graphs have irreducible characteristic polynomials. Haemers has conjectured that the proportion of graphs on $n$ vertices that are determined by their characteristic polynomials goes to 0 as $n\to\infty$. My suspicion is that pairs of cospectral graphs are not normally integrally equivalent. I am hoping that if I learn more number theory, I might be able to confirm this.

  • $\begingroup$ For item (1), why not work through a proof for the case of maximal orders yourself to get the proof to work for non-maximal orders? It certainly is true that the number of ideal classes in a non-maximal order is finite, but instead of trying to show every ideal class contains an integral ideal with index below some bound, show every ideal class contains a fractional ideal that contains the order with index below some bound. There's a step in the usual finiteness proof where you invert ideals, but just avoid taking that step. $\endgroup$ – KConrad Aug 6 '11 at 10:29
  • $\begingroup$ KConrad: Thanks. I will follow your advice. I gave one reason why I wanted a reference. Another is that if I ever wrote down anything about this for graph theorists and tried to publish, a referee could well insist on a reference. Also, I suspect it's your online notes I have been studying. Thanks for those, they're very nice. $\endgroup$ – Chris Godsil Aug 6 '11 at 12:57
  • 1
    $\begingroup$ 1. Since these rings are not Dedekind, ideal class group means class group of ideals coprime to the index. And this is finite since you can realize it as a (quotient of a) suitable ray class group. In the case of quadratic extensions, these class groups actually were called "ring class groups" way back then. 2. I'm not aware of any progress beyond what can be found in Cohn's appendix by Taussky. 3. No. $\endgroup$ – Franz Lemmermeyer Aug 6 '11 at 12:59
  • $\begingroup$ Chris, the finiteness of the number of ideal classes (not just invertible ideal classes) in non-maximal orders in not in any of those notes on my website. $\endgroup$ – KConrad Aug 6 '11 at 17:33
  • $\begingroup$ For 1: in case you can't find anything else, and if you can read French, you can find a proof as a homework problem at agreg.org/sujets/MG11.pdf. This was the topic of the French agrégation exam in 2011. $\endgroup$ – Oblomov Nov 15 '12 at 13:24

A very belated answer to 1), but: I just saw that this is treated very nicely in Curtis and Reiner's Representation Theory of Finite Groups and Associative Algebras. Theorem 20.6 therein not only works with non-maximal orders $\mathbb{Z}[\theta]$ but even does the non-commutative case as well. And the argument is very clean and simple.

2) is a nice question by the way, and one that I have wondered about over the years. As Franz says, you need to be careful about what you mean by the "ideal class group" in this case. For any domain $R$ the quotient of the monoid of the nonzero ideals of $R$ by the submonoid of nonzero principal ideals gives you a monoid, say $H(R)$: let's call it the ideal class monoid. This is the monoid that your question 1) asks for a proof of the finiteness of when $R$ is an order in a number field $K$. (Well, your question asks about the monogenic case, but the proof of course doesn't need that.) It is known that $H(R)$ is a group iff $R$ is a Dedekind domain, so in the arithmetic case iff $R$ is integrally closed in its fraction field. In every other case there are non-invertible ideal classes, so to get a group you should pass to the group of units of the monoid $H(R)$: by definition this group is the Picard group of $R$ (it is also the group of isomorphism classes of rank one projective $R$-modules under tensor product, or if you like of isomorphism classes of line bundles on the scheme $\operatorname{Spec} R$). If e.g. $R$ is Noetherian, integrally closed, there is also a divisor class group $\operatorname{Cl}(R)$ whose nonvanishing is the obstruction to $R$ being a UFD. There is a canonical injection $\operatorname{Pic}(R) \hookrightarrow \operatorname{Cl}(R)$ which need not be an isomorphism: the most famous example is probably $\mathbb{C}[x,y,z]/(xy-z^2)$ which has two-element divisor class group but trivial Picard group. The map is an isomorphism if $R$ is "locally factorial", so e.g. if $R$ is nonsingular, hence always in dimension one.

The one example I know of $H(R)$ for a nonmaximal order is $R = \mathbb{Z}[\sqrt{-3}]$, where the monoid $H(R)$ has order $2$. There is up to isomorphism exactly one order $2$ commutative monoid which is not a group: the nonidentity element $\bullet$ must be "absorbing": $x \bullet = \bullet$ for all $x$. In this case a representative for the absorbing element is the prime ideal $\mathfrak{p} = \langle 2, 1+ \sqrt{-3} \rangle$. This is the unique non-principal prime ideal in $R$ (something which cannot happen in a Dedekind domain!). I would be interested to know if anything whatsoever is known or conjectured about which finite non-cancellative monoids arise as $H(R)$ as $R$ ranges over orders in algebraic number fields.

If $R$ is a nonmaximal order in a number field $K$ with maximal order $\tilde{R}$, there is a nonzero ideal $\mathfrak{f}$, the conductor, which is the set of all $x \in \tilde{R}$ such that $\tilde{R} x \in R$. This is an ideal in both $\tilde{R}$ and $R$, and in fact it is the largest ideal of $\tilde{R}$ which is also an ideal of $R$. The primes $\mathfrak{p}$ of $R$ containing $\mathfrak{f}$ are precisely the singular points on $\operatorname{Spec} R$. Every ideal prime to $\mathfrak{f}$ is invertible and factors uniquely into primes, and every invertible ideal can be adjusted in its equivalence class to be prime to $\mathfrak{f}$. Thus the noninvertible ideals are the ones with support which cannot be "moved off of $\mathfrak{f}$". This material can be found in Neukirch's Algebraic Number Theory.


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