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Let $\theta = \sqrt[3]{2}$, and $M = \{4, \theta, \theta^2\}$, $N = \{1, 2\theta, 2\theta^2\}$ be finitely generated $\mathbb{Z}$-modules in $\mathbb{Q}(\theta)$. I would like to show either $M \cong N$ (which seems to be false) or $M \not\cong N$, but I do not know how to get started. Can anyone help, maybe by providing me with some invariants of isomorphism of this sort? Thanks in advance!

edit: I am sorry for the poor quality of my question. Thanks to Mr. julio_es_sui_glace, whose answer has solved my question. Still I would like to provide some contexts so that the description may be improved.

I comes up with the problem when reading 2.2, Chapter 2 of Borevich & Shafarevich's Number Theory. Let $M$ denote a finite-generated, torsion-free and full $\mathbb{Z}$-module in an algebraic number field, $\mathfrak{O}_M$ its coefficient ring. The book constructs first a injection $\mathfrak{O}_M \to M$, and shows that they are of the same rank. I thought at that time this was sufficient for $\mathfrak{O}_M \cong M$ (module isomorphism). But later on the book mentions another injective isomorphism $M \to \mathfrak{O}_M$, which made me doubt if $\mathfrak{O}_M \cong M$. To understand the necessity of building the latter isomorphism, I tried to calculate some examples, and (as glace said) confused module isomorphism with ring isomorphism.

Now I guess the book constructs the latter isomorphism because it is explicit (by multiplying $M$ with a certain natural number), showing not only the coincidence of module structure, but also that $M$ is similar to a module in $\mathfrak{O}_M$ which is used in further developments (e.g. showing that $M$ contains only a finite number ofpairwise-nonassociate elements with given norm).

I should thank those who give me advice to improve the question. I also hope there will be answers improving my understanding on the intention of the two isomorphism constructed in the book.

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You have a clear isomorphism of modules (but not rings) given by any bijection between $\{4,\theta,\theta^2\}$ and $\{1,2\theta,2\theta^2\}$.
This is clear since these modules are free since both $\{4,\theta,\theta^2\}$ and $\{1,2\theta,2\theta^2\}$ are basis of the $\mathbb{Q}$-vector space $\mathbb{Q}[\theta]$.
Free modules of the same rank are isomorphic!

e.g. you have the isomorphism of modules given by $$\begin{array}{rcl} M & \longrightarrow & N\\ 4 & \longmapsto& 1 \\ \theta & \longmapsto& 2\theta\\ \theta^2 & \longmapsto& 2\theta^2 \end{array}$$

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