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, the above isomorphism holds by 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 $\mathbb{Q}$.


As Mohan already observed, the Smith Norm Formal Theorem can be extended in your sense to the class of Dedekind domains. 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 of a 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 is one instance because of the principal ideal theorem, see also this MO post for further examples.

Edit. It turns out that the isomorphism holds more generally for Prüfer domains 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.


This is false for general rings, but for Dedekind domains, it is alright. Use the fact that everything localizes well and they are supported only at finitely many prime ideals, containing $I_m$, $m=\mathrm{rk} A$. So, you may semi-localize and then you are in the pid situation.

  • $\begingroup$ Thus, this holds for the ring of integers in a quadratic extension of $\mathbb{Q}$. Do you have any counterexample for an order $R$ in $\mathbb{Q}(\sqrt{-n})$? $\endgroup$ – Roberto Pagaria Jul 6 '18 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.