Return to Answer

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 $\mathfrak{m}$ [1, 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.

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

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.

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 $\mathfrak{m}$ such that the $\mathfrak{m}R_{\mathfrak{m}}$ cannot be generated by a single element as an ideal of the localization $R_{\mathfrak{m}}$ at $\mathfrak{m}$ [1, 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.

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

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 $\mathfrak{m}$ [1, 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.

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

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 $\mathfrak{m}$ such that the $\mathfrak{m}R_{\mathfrak{m}}$ cannot be generated by a single element as an ideal of the localization $R_{\mathfrak{m}}$ at $\mathfrak{m}$ [1, 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.

[1] H. Matsumura, "Commutative ring theory", 1986.

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 $\mathfrak{m}$ such that the $\mathfrak{m}R_{\mathfrak{m}}$ cannot be generated by a single element as an ideal of the localization $R_{\mathfrak{m}}$ at $\mathfrak{m}$ [1, 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.

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