Let $R$ be a (commutative) domain and let $Q$ be its fraction field.
Consider a morphism $f\colon R^n \to R^m$, i.e. a matrix $A \in M(m,n;R)$, and let $K= \operatorname{coker} f$.


Let $I\_k=(\det \operatorname{min}\_k(A))$ be the ideal of $R$ generated by all determinants of minors of $A$ of size $k$.
I wonder if 
\begin{equation}
\operatorname{Tor}\_1^R(K,Q/R) \simeq \prod\_{k=1}^{\operatorname{rk} A} \frac{I\_{k-1}}{I\_{k}}.
\end{equation}
If $R$ is a [PID](https://en.wikipedia.org/wiki/Principal_ideal_ring), the above isomorphism holds by [Smith Normal Form](https://en.wikipedia.org/wiki/Smith_normal_form).


Does it hold for other classes of rings?


I'm particularily interested in the case of $R$ is an order in a quadratic imaginary extension of $\mathbb{Q}$.



Generalized Smith Theorem for the torsion of cokernels
As Mohan already observed, the [Smith Norm Formal Theorem](https://en.wikipedia.org/wiki/Smith_normal_form) can be extended in your sense to the class of [Dedekind domains](https://en.wikipedia.org/wiki/Dedekind_domain). This is somehow the best we can get:



> 
> 
> > 
> > **Claim.** Let $R$ be a Noetherian domain which is not a Dedekind domain. Then there is a maximal ideal $\mathfrak{m}$ of $R$ such that the dimension of $\mathfrak{m}/\mathfrak{m}^2$ over $R/\mathfrak{m}$ is at least $2$. In particular, $R/\mathfrak{m} \times R/\mathfrak{m}$ and $R/\mathfrak{m} \times \mathfrak{m}/\mathfrak{m}^2$ are not isomorphic as $R$-modules although we have $I\_0 = R, I\_1 = \mathfrak{m}$ and $I\_2 = \mathfrak{m}^2$ for an inclusion map $\mathfrak{m} \times \mathfrak{m} \subset R \times R$.
> > 
> > 
> > 
> 
> 
> 


By an inclusion map, I mean any $R$-module homomorphism $f: R^n \rightarrow R^2$ with image $\mathfrak{m} \times \mathfrak{m}$.



> 
> 
> > 
> > *Proof.* By hypothesis, we can find a maximal ideal $\mathfrak{m}$ such that $\mathfrak{m}R\_{\mathfrak{m}}$ cannot be generated by a single element as an ideal of the localization $R\_{\mathfrak{m}}$ at $\mathfrak{m}$ [3, Theorem 11.2]. Thus the dimension of $\mathfrak{m}R\_{\mathfrak{m}}/\mathfrak{m}^2R\_{\mathfrak{m}}$ over $R/\mathfrak{m}$ is at least two. As we have $\mathfrak{m}/\mathfrak{m}^2 \simeq (\mathfrak{m}/\mathfrak{m}^2)\_{\mathfrak{m}} \simeq \mathfrak{m}R\_{\mathfrak{m}}/\mathfrak{m}^2R\_{\mathfrak{m}}$, the result follows.
> > 
> > 
> > 
> 
> 
> 


As a result, the Smith Normal Form Theorem can be extended in your sense to an [order](https://en.wikipedia.org/wiki/Order_(ring_theory)) of a [number field](https://en.wikipedia.org/wiki/Algebraic_number_field) if 
and only if this order is maximal.


Let us conclude with a remark on two obvious *generalizations*.


If $R$ is any commutative ring with identity, OP's isomorphism trivially holds whenever $m = 1$. The existence of a Smith Normal Form also trivially guarantees the existence of OP's isomorphism. By definition, every *Elementary Divisor Ring (EDR)* ensure this existence, independently of $m$ and $n$, and there are EDRs which aren't principal ideal rings. The [ring of algebraic integers](https://en.wikipedia.org/wiki/Algebraic_integer) is one instance because of the [principal ideal theorem](https://en.wikipedia.org/wiki/Principal_ideal_theorem), see also this [MO post](https://mathoverflow.net/questions/31275/does-smith-normal-form-imply-pid/31306#31306) for further examples.


**Edit.** It turns out that the isomorphism holds more generally for [Prüfer domains](https://en.wikipedia.org/wiki/Pr%C3%BCfer_domain) of finite character, i.e., Prüfer domains in which every nonzero element is contained in only finitely many maximal ideals. This is [2, Proposition 1].
For an example of one of these rings that is neither Bézout nor
Noetherian, see [1].




---


[1] W. Heinzer, "Quotient overrings of integral domains", Mathemafika 17, 1970  

[2] L. Levy, "Invariant Factor Theorem for Prüfer Domains
of Finite Character ", 1985.  

[3] H. Matsumura, "Commutative ring theory", 1989.


