I am looking for a reference to the following well known fact:
Any finitely generated torsion free module over a discrete valuation ring is free of finite rank.
Finitely generated modules over discrete valuation ring.
algebraic-number-theorycommutative-algebra
Related Solutions
The answer is that $f$, for certain rings, cannot be onto $M$, so this line of reasoning isn't going to work. If, for example $M=R^n$ $n>1$, and $R$ is finite, then it's going to be impossible for $R$ to map onto $R^n$.
It is commendable that you had this idea that "$R$ is Noetherian, so let me find a map to $M$ from $R$..."! If we revise your idea slightly, you'll be on the right track.
The thing to notice is that $R^n$ is a Noetherian module if $R$ is Noetherian. After you know that, you can cover $M$ more completely than you could with simply $R$. Then proceed with your "covering $M$" idea.
Will be waiting if you have more questions... good luck!
As Mariano has pointed out in the comments, over a local ring, projective and free are equivalent, so you are asking whether every finitely generated torsion-free module over a local domain is free.
This is not the case. Indeed, first observe that over any commutative ring $A$, an ideal $I$ is free if and only if it is principal and torsion-free. If $I$ is principal and torsion-free, then the singleton consisting of any generator is a basis for $I$. On the other hand, note that any two distinct elements $x, y \in I$ are linearly dependent, because $y \cdot x - x \cdot y = 0$ is a nontrivial $A$-linear relation.
Hence, it suffices to give an example of a local domain $A$ with an ideal $I$ that is not principal, since $I$ will be torsion-free but not free. Mariano already suggested an example of such a ring in the comments: for $F$ a field, consider the ring $A := F[X, Y]_{\langle X, Y \rangle}$. Then $A$ is a local ring with maximal ideal $M := \langle X, Y \rangle$. But $M$ cannot be generated by a single element; there are many proofs of this, but one is to observe that $M/M^{2}$ is an $A/M \cong F$-vector space of dimension $2$, and so $M$ cannot be principal or else $M/M^{2}$ would be a $1$-dimensional $F$-vector space.
I think it's worth noting that this kind of example is quite common. If $A$ is a Noetherian local ring with unique maximal ideal $M$, then by Krull's principal ideal theorem, the minimum number of generators of $M$ as an $A$-module is bounded below by the Krull dimension of $A$. Examples of local domains with Krull dimension greater than one abound, and give rise to an example in the same way highlighted above. The particular example we gave belongs to an important class of rings you will encounter when studying commutative algebra, namely so-called regular local rings. A Noetherian local ring $A$ with unique maximal ideal $M$ is a regular local ring if the minimal number of generators of $M$ as an $A$-module is equal to the Krull dimension of $A$.
Best Answer
A discrete valuation ring is a principal ideal domain and a finitely generated torsion free module over a PID is free of finite rank. This in turn follows from the structure theorem for modules over a PID.