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:

## Questions

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.]

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.]

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.

containsthe 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