I'm reading Cohen-Macaulay Rings by Bruns & Herzog. The lemma below which is used in the proof of the Proposition 1.4.9 and whose proof is left to the reader is difficult for me to prove:
Lemma.
Let $R$ be a Noetherian ring, and $M$ be a finitely generated $R$-module. Then $M$ is a projective module and has rank $r$ (i.e. $M\otimes Q$ where $Q$ is the total ring of fractions of $R$ is a free $Q$-module of rank $r$) if and only if $M_{\mathfrak{p}}$ is a free $R_{\mathfrak{p}}$-module of rank $r$.
To prove this, the lemmas below may be useful:
Proposition 1.4.3.
Let $R$ be a Noetherian ring, and $M$ an $R$-module with a finite free presentation $F_{1}\rightarrow F_{0}\rightarrow M\rightarrow 0$. Then the following are true:
(a) $M$ has rank $r$;
(b) for all prime ideals $\mathfrak{p}\in \mathop{\mathrm{Ass}}R$ the $R_{\mathfrak{p}}$-module $M_{\mathfrak{p}}$ is free of rank $r$.
Theorem 7.12. of Commutative Algebra by H. Matsumura (Japanese version).
Let $R$ be a ring and $M$ be an $R$-module of finite presentation. Then $M$ is projective if and only if $M_{\mathfrak{m}}$ is a free $R_{\mathfrak{m}}$-module for all maximal ideals $\mathfrak{m}\subseteq R$.
My idea for the proof.
By theorem 7.12, we find that $M$ is projective iff $M_{\mathfrak{m}}$ is free for all maximal ideals iff $M_{\mathfrak{p}}$ is free for all primes. Therefore the sufficiency follows from theorem 1.4.3. For the necessity, it follows that $M_{\mathfrak{p}}$ is free of rank $r$ for all $\mathfrak{p}\in\mathop{\mathrm{Ass}}R$. However, we don't know anything about those primes which are not associated primes. How can we manage it?
If $q$ is an arbitrary prime ideal, then it contains a minimal prime, say $p$. We know that $M_q$ is a free $R_q$-module of finite rank, say $s$. Then any localization of $M_q$ is a free module of rank $s$, too. But $(M_q)_{pR_q}\simeq M_p$, so its rank is actually $r$. (Recall that a minimal prime is associated.)