Let $A$ be a [UFD](https://en.wikipedia.org/wiki/Unique_factorization_domain), $M$ a free module over $A$, and $N$ a finitely generated rank 1 submodule. I have a relatively quick proof (about half a page) that if $(A\smallsetminus\{0\})^{-1}N \cap M = N$, then $N$ is also free. This is done by induction, reducing the number of generators to 1. 


However, I feel like my proof is superficial, and that there should be a "deeper" proof in some sense. So my question is:


Is there some deep reason that the above statement should be obvious?



UFD free modules of rank 1
Let $A$ be an integral domain and let $N \subseteq M$ be two modules over $A$. The condition $S^{-1}N \cap M = N$ for $S = A \setminus \{0\}$ is equivalent to $N \cap aM = aN$ for every $a \in A$, provided that $M$ is torsion-free. Hence, in this case, $N$ is a pure sub-module of $M$ in a weak sense.


We can prove the following claim without using induction on the number of generators of the sub-module $N$.



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> > **Claim.** Let $A$ be a a [GCD domain](https://en.wikipedia.org/wiki/GCD_domain), let $M$ be a free module over $A$ and let $N$ be a finitely generated sub-module of $M$ of rank $1$. If 
> >  $N \cap aM = aN$ for every $a \in A$ then $N \simeq A$.
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The following lemma will help us reduce to the case $M = A$.



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> > **Lemma 1.** Let $A$ be a [GCD domain](https://en.wikipedia.org/wiki/GCD_domain), and let $M$ be a free module over $A$ of finite rank $n > 0$. Let $x$ be a non-zero element in $M \otimes\_A K \simeq K^n$ where $K$ denotes the field of fractions of $A$. Then $M \cap Ax \simeq A$.
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> > *Proof.* Write $x = (\frac{a\_1}{b\_1},\dots, \frac{a\_n}{b\_n})$, with $\text{gcd}(a\_i, b\_i) = 1$ for every $i$. For $a \in A$, we have $ax \in A^n$ if and only if $a \in \bigcap\_i Ab\_i$. Since $A$ is a GCD domain, the latter intersection is a principal ideal.
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Now we can proceed to the proof of the claim.



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> > *Proof of the Claim.* As $N$ is finitely generated, we can assume without loss of generality, that $M$ has finite rank $n > 0$. We identify then $M$ and $N$ with their images in $K^n = M \otimes\_A K$ where $K$ is the field of fractions of $A$. Since the rank of $N$ is $1$, there is $x \in K^n$ such that $N \subseteq Kx$. As $N$ is finitely generated, we can assume that $N \subseteq Ax$. Thus we can identify both $N$ and $M \cap Ax$ with ideals of $A$. Because of Lemma 1, we can assume moreover that $M = A$. Given $a \in N \setminus \{0 \}$, we have $N \cap Aa = Aa \subseteq aN$. Therefore $N = A$.
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