Let $A$ be a commutative ring with 1, and let $P$ be an $A$-module. According to Bourbaki (Algèbre commutative, Ch.II, $\S 5$, No 2, Thm 1), we have the following equivalence:
$(1)$ $P$ is finitely generated and projective
$(2)$ there exists $s_1,\ldots,s_n\in A$ such that $(s_1,\ldots,s_n)=A$ and $P_{s_i}$ is a free $A_{s_i}$-module of finite rank.
I would like to find a direct proof of this equivalence, without using too much intermediate results
I know how to prove it when I replace $(2)$ by
$(2')$ $P$ is finitely presented and there exists $s_1,\ldots,s_n\in A$ such that $(s_1,\ldots,s_n)=A$ and $P_{s_i}$ is a free $A_{s_i}$-module of finite rank.
Before asking my questions, let me sketch my proof of $(2')\Rightarrow (1)$.
We prove that for any surjective linear map $u:M\to N$, then $u_*:Hom(P,M)\to Hom(P,N)$ is surjective. For, it suffices to prove that the localizations $(u_{s_i})_*$ is surjective are surjective.
Now since $P$ is finitely presented, the canonical map $Hom(P,M)_{s_i}\to Hom(P_{s_i},M_{s_i})$ is bijective, and same replacing $M$ by $N$. Now, since $u$ is surjective, $(u_{s_i})_*$ is surjective. Then, use freeness of $P_{s_i}$ + commutativity of the diagram you think to get surjectivity of $(u_*)_{s_i}$ .
Questions.
(i) Does condition (2) implies that $P$ is finitely presented ? If yes, how to prove it ?
(ii) If not, do you know a direct proof of $(2)\Rightarrow (1)$ which does not involve to much intermediate rsults (in particular, i would like to avoid too many results on faithfull flatness)
Thanks for your help.