Let $M$ be a module over a ring or an algebra $A$.I have seen three definitions of finitely presented modules:
(1) A module $M$ is called finitely presented if there is an exact sequence of $A$-modules: $0 \rightarrow L \rightarrow F \rightarrow M \rightarrow 0$, where $F$ is a free module of finite rank, $L$ is finitely generated;
(2) A module $M$ is called finitely presented if there is an exact sequence of $A$-modules: $F_2 \rightarrow F_1 \rightarrow M \rightarrow 0$, with $F_1,F_2$ free modules of finite rank;
(3) A module $M$ is called finitely presented if there is an exact sequence of $A$-modules: $P_2 \rightarrow P_1 \rightarrow M \rightarrow 0$, with $P_1,P_2$ finitely generated projetive modules.
I can get (2) by (1), get (3) by (2). But I can't get they are equivalent. So who can tell me how to prove they are equivalent?
Suppose $$P_2\to P_1\to A\to 0$$ is as in (3). Since $P_1$ is finitely generated and projective, there is a finitely generated module $Q$ such that $P_1\oplus Q=F$ is a finite rank free module. Now take the direct sum of our exact sequence and the exact sequence $$Q\stackrel{1_Q}\to Q\to 0\to 0$$ to get another exact sequence $$P_2\oplus Q\stackrel{f}\to F\stackrel{g}\to A\to 0.$$ Since $P_2$ and $Q$ are finitely generated, $\operatorname{im}(f)=\ker(g)$ is finitely generated (it is generated by the images of generators of $P_2$ and $Q$ under $f$). Let $L=\ker(g)$. We then have a short exact sequence $$0\to L\to F\stackrel{g}\to A\to 0$$ which satisfies the requirements of (1).