Do we have $\mathfrak{a}\mathfrak{c}/\mathfrak{b}\mathfrak{c} \cong \mathfrak{a}/\mathfrak{b}$ for fractional ideals in an order of a number field?

Question

Let $\mathcal{O}$ be an order in a number field $K$ and let $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}$ be fractional ideals of $\mathcal{O}$. If $\mathcal{O}$ is a Dedekind domain, then this answer proves that there is an isomorphism $$ \mathfrak{a} \mathfrak{c}/\mathfrak{b}\mathfrak{c} \cong \mathfrak{a}/\mathfrak{b}. $$ My questions are:

1. Does this isomorphism still hold if $\mathcal{O}$ is not a Dedekind domain? The proof in the linked answer makes lots of use of invertibility of fractional ideals, and the Lemma it states uses unique factorisation into prime ideals, so I haven't been able to generalise the proof. However, it seems plausible that these properties are convenient but not essential.
2. If not, then is there some salvage? I am particularly interested in the size of these quotients. It would solve my problem if we had something like $[\mathfrak{a} \mathfrak{c}:\mathfrak{b}\mathfrak{c}] \leq [\mathfrak{a}:\mathfrak{b}]$. Now that I have written this in terms of indices of modules, I am strongly reminded of the statement that $[LM : M] \leq [L : K]$ for finite field extensions $L/K$ and $M/K$ inside some common extension. I don't see a way to prove $[\mathfrak{a} \mathfrak{c}:\mathfrak{b}\mathfrak{c}] \leq [\mathfrak{a}:\mathfrak{b}]$ in a similar way, but the superficial similarity gives me hope that it might be true.

Answer 1

If $\mathfrak c$ is an invertible ideal in $\mathcal O$, then $\mathfrak a\mathfrak c/\mathfrak b\mathfrak c \cong \mathfrak a/\mathfrak b$ as $\mathcal O$-modules. However, in general that isomorphism is not true in a non-maximal order. In fact, in general we can't have $|\mathfrak a\mathfrak c/\mathfrak b\mathfrak c| = |\mathfrak a/\mathfrak b|$ for all nonzero ideals $\mathfrak a$, $\mathfrak b$, and $\mathfrak c$ in $\mathcal O$. More specifically, taking $\mathfrak a = \mathcal O$, we can't have $|\mathfrak c/\mathfrak b\mathfrak c| = |\mathcal O/\mathfrak b|$ for all nonzero ideals $\mathfrak b$ and $\mathfrak c$ in $\mathcal O$. (All these indices are finite since all nonzero ideals in an order have finite index.)

Indeed, suppose $|\mathfrak c/\mathfrak b\mathfrak c| = |\mathcal O/\mathfrak b|$ for all nonzero ideals $\mathfrak b$ and $\mathfrak c$ in $\mathcal O$. Then $$ [\mathcal O:\mathfrak b\mathfrak c] = [\mathcal O:\mathfrak c][\mathfrak c:\mathfrak b\mathfrak c] = [\mathcal O:\mathfrak c][\mathcal O:\mathfrak b], $$ so the counting function ${\rm N}(\mathfrak a) := [\mathcal O:\mathfrak a]$ on nonzero ideals $\mathfrak a$ in $\mathcal O$ is totally multiplicative. And that implies $\mathcal O$ is a Dedekind domain, by a theorem of Butts and Wade. (Theorem 2 in "Two criteria for Dedekind domains", Amer. Math. Monthly 73 (1966), 14-21) here.

Your wish that $[\mathfrak a\mathfrak c:\mathfrak b\mathfrak c] \leq [\mathfrak a:\mathfrak b]$ also has many counterexamples. Taking $\mathfrak a = \mathcal O$ and $\mathfrak b = \mathfrak c$, you're in particular wishing that $[\mathfrak c:\mathfrak c^2] \leq [\mathcal O:\mathfrak c]$, but below are many examples where $[\mathfrak c:\mathfrak c^2] > [\mathcal O:\mathfrak c]$.

For an integer $c \geq 2$ and a number field $K$ of degree $n > 1$ over $\mathbf Q$, set $\mathcal O = \mathbf Z + c\mathcal O_K$. This is an order in $K$. Set $\mathfrak c := c\mathcal O_K$, which is an ideal in $\mathcal O$. Then $$ \mathcal O/\mathfrak c = (\mathbf Z + \mathfrak c)/\mathfrak c \cong \mathbf Z/(\mathbf Z \cap \mathfrak c) = \mathbf Z/(\mathbf Z \cap c\mathcal O_K) = \mathbf Z/c\mathbf Z, $$ so $[\mathcal O:\mathfrak c] = c$. Also $\mathfrak c^2 = c^2\mathcal O_K$, so $$ \mathfrak c/\mathfrak c^2 = c\mathcal O_K/c^2\mathcal O_K \cong \mathcal O_K/c\mathcal O_K \cong (\mathbf Z/c\mathbf Z)^n $$ as abelian groups. Thus $[\mathfrak c:\mathfrak c^2] = c^n > c = [\mathcal O:\mathfrak c]$. Taking for $K$ a number field of very large degree $n$ over $\mathbf Q$, we get examples where the ratio $[\mathfrak c:\mathfrak c^2]/[\mathcal O:\mathfrak c] = c^{n-1} \geq 2^{n-1}$ gets very large.