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I am stuck at proving

Every $R$-module $M$ has a maximal linearly independent subset $A$. If the submodule $M_A$ is generated by $A$, then $M_A$ is free on $A$, and for a non-zero submodule $M_1\subseteq M$, $M_1 \cap M_A \neq 0$.

I know for the first part I should just apply Zorn's lemma, but I am not clear about the rest of the problem. Could anyone help me with that?

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As you suggested, the existence of a maximal linearly independent set is a fairly standard use of Zorn's lemma.

To show that $M_A$ is free, you need an isomorphism $M_A\to \bigoplus_{\lambda\in\Lambda} Re_\lambda$ for some index set $\Lambda$ and where the $e_\lambda$ are formal symbols. The natural choice is to let $\Lambda=A$ and send $a\mapsto e_a$. Extend this by $R$-linearity to show it is a homomorphism; surjectivity is easy, and injectivity follows from linear independence.

The last statement isn't true, and I don't see an easy fix. As a counterexample, let $R=\Bbb Z$ and $M=\Bbb Z_2$. Then $M$ does not have any nonempty linearly independent subsets, so $M_A=0$. This is not just a problem with the empty set or the zero module; we can just as easily consider $M=\Bbb Z\oplus \Bbb Z_2$, where a qualitatively similar issue arises with $A=\{(1,0)\}$.

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