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I have read that the $\mathbb Z$-module $(\mathbb Q,+)$ is not isomorphic to a submodule of $\mathbb Z^n$, for any $n\in \mathbb Z^+$. I come up with a proof (below) that seems a bit facticious, and likely, there is a cleaner way to do it. What do you think?

Consider the set $X:=\{\frac 1n:n\in \mathbb Z^+\}\subset \mathbb Q$, whose cardinality is not finite. For any $x\in X$, there is some $r\in\mathbb Z$ satisfying $r\cdot x=1$. Hence if $\mathbb Q$ is injected in $\mathbb Z^n$, and the image of $1$ is $(u_1,\dots,u_n)$, there should be infinitely many $n$-tuples $(a_1,\dots,a_n)$ such that $r\cdot (a_1,\dots,a_n)=(u_1,\dots,u_n)$ for some $r\in \mathbb Z$. The $n$-tuples in question are not infinitely many however (they are as many as the divisors of the MCM of $a_1,\dots, a_n$).

CRinge
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    Proof looks correct to me! You could alternatively try to show it's not free since any submodule of $\mathbb{Z}^n$ is free (take a rank $1$ submodule of $\mathbb{Q}$ and show that nothing generates it) – Sofía Marlasca Aparicio Aug 07 '22 at 11:37

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