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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?

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  • $\begingroup$ In the first definition, isn't there a term missing in the exact sequence? The last morphism should be surjective. $\endgroup$ – Pierre-Guy Plamondon Sep 29 '16 at 16:20
  • $\begingroup$ Yes, I have edited it. Thank you for your advice. $\endgroup$ – Daisy Sep 30 '16 at 1:29
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    $\begingroup$ To Go from 3 to 1 just replace $P_2$ by its image under the map to $P_1$. $\endgroup$ – Mariano Suárez-Álvarez Sep 30 '16 at 1:31
  • $\begingroup$ @Mariano Suárez-Álvarez♦ But condition (1) needs the middle term to be a free module of finite rank, and L to be finitely generated. I know a projective module can be completed to a free module, but I can't make sure the free module is of finite rank and L to be finitely generated. $\endgroup$ – Daisy Sep 30 '16 at 1:40
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    $\begingroup$ Just add to the middle term whatever you need to make it free (and a copy of that on the third term) $\endgroup$ – Mariano Suárez-Álvarez Sep 30 '16 at 1:47
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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).

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