Let $M$ be a finitely generated module over a Noetherian domain $R$. Further, let $\mathfrak p$ be a prime ideal of $R$. Then by NAK all the minimal generating sets of the $R_\mathfrak p$-module $M_\mathfrak p$ have the same cardinality, which equals the dimension of the $R_\mathfrak p/\mathfrak pR_\mathfrak p$-vector space $M_\mathfrak p/\mathfrak pR_\mathfrak p M_\mathfrak p$, where $\mathfrak pR_\mathfrak p$ is the maximal ideal in the local ring $R_\mathfrak p$.
How does one show that $M$ is flat iff the cardinality of minimal generating sets of $M_\mathfrak p$ is independent of $\mathfrak p$?
There is a proof of this (or something similar) in P. Clark's lecture notes (in fact, it is left as an exersise), but it uses a few theorems on projective modules. I'm wondering whether there is a way to establish this without appealing to projective modules?
(Note $M$ is flat iff $M_\mathfrak p$ is flat for all primes $\mathfrak p$ in $R$ iff $M_\mathfrak p$ is free for all primes $\mathfrak p$ in $R$ iff there are $r_1,\dots,r_n\in R$ such that $R=(r_1,\dots,r_n)$ and $M_{r_i}$ are all free over $R_{r_i}$.)
The forward direction is a formal consequence of one of the equivalent definitions:
The characterization
shows that the rank of $M$ at all primes in $V(\mathfrak p)$ is the same as the rank at $\mathfrak p$. Since $R$ is a domain, we can pick $\mathfrak p=0$ and are done.
The backward direction can be shown like this:
Let $n$ be the dimension of $M_{(0)}$ over the fraction field. Let $\mathfrak p$ be a prime ideal. By assumption, we know that $M_{\mathfrak p}$ can be generated by $n$ elements. A standard argument in localization shows that $M_{(0)}$ can be generated by the same $n$ elements. Since it is a vector space of dimension $n$, the elements form a basis, hence they are necessarily $R_{(0)}$-linear independent. In particular they are $R_{\mathfrak p}$-linear independent, hence they form a basis of $M_{\mathfrak p}$.