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](https://mathoverflow.net/questions/52461 "Title"), 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:


Questions
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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?


### Remark


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.



Ideal classes and integral similarity
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*.


