A module $M$ is said to be a square-free module if for any two submodules $A,B \leq M$ with $A \cap B = 0$ and $A \cong B$ we have that $A=B=0$. I claim that $\mathbb{Q}$ is a square-free as a left $\mathbb{Z}$-module. Does anybody know how to prove that fact?
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2try to show that any two non-zero subgroups of $\mathbb Q$ have a nontrivial intestection – user8268 Sep 15 '21 at 18:13
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Any nontrivial $\mathbb{Z}$-submodule $A$ of $\mathbb{Q}$ contains a nonzero integer, since if $x = a/k$ is a nonzero element of $A$ with $a,k \in \mathbb{Z}$, then $a = kx$ is a nonzero integer in $A$.
Let $a,b$ be nonzero integers in nontrivial $\mathbb{Z}$-submodules $A$, $B$, respectively. Then $ab$ is a nonzero element in $A\cap B$.
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