UFD free modules of rank 1 Let $A$ be a UFD, $M$ 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?
 A: 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 $N$.


Claim. Let $A$ be a a  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 $N \simeq A$.


The following lemma will help us reduce to the case $M = A$.


Lemma 1. Let $A$ be a  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 $M \cap Ax \simeq A$.
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.


Now we can proceed to the proof of the claim.


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 $N = A$.


A: Free modules $M$ over a GCD domain have the following property: If $x \in M$, $a^{-1} x \in M$, and $b^{-1} x \in M$, and $a$ and $b$ have no common factors, then $(ab)^{-1} x \in M$. Indeed, it suffices to check this for the free module of rank one, where the claim is that if $x$ is divisible by $a$ and $b$ where $a$ and $b$ share no common factors then $x$ is divisible by $ab$ - obvious from unique factorisation.
If $N \subseteq M$, $M$ has this property, and $N$ is closed under division inside $M$, then $N$ shares this property.
Any rank one, finitely generated, module over a Noetherian ring with this property is free: Let $x$ generate a maximal free submodule, $y$ another element, then because it is rank one, $y = (a/b)x$ for some $a,b$ with no common factors, hence $ax=by$ is divisible by $a$ and $b$, hence $x$ is divisible by $b$. Since $x/b$ cannot generate a larger free submodule, $b$ is a unit, and hence $y$ is a multiple of $x$.
