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For rings with unit there are at least three ways to define a projective module:

  1. The universal property, i.e. $P$ is projective if for any epimorphism $M\to N$ and any morphism $P\to N$ there exists a morphism $M\to P$ such that the diagram commutes.
  2. A projective module is a direct summand of a free module.
  3. Any epimorphism onto $P$ splits.

My question is what relation there is between these three conditions when the rings does not have a unit. Of course 1. implies 3., but what about the other directions? I'm most interested in "3. implies 1.?".

Edit: @t.b. has answered the 3. implies 1. In searching the web I found the claim that not even the rng (as the most basic free module) itself is projective. (see e.g. Anh, Marki: Morita equivalence over rings without identity and Arando Pino, Rangaswamy, Siles Molina: Weakly regular and self-injective Leavitt path algebras over arbitrary graphs.) So 2. seems not to imply 1. However I didn't find a counterexample in those papers. So, what is a counterexample for that?

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For 3. implies 1. take the pull-back $Q = P \mathop{\times_N} M$ of $P \to N \gets M$, then choose a splitting of $Q\to P$ and observe that $P \to Q \to M$ is a lift of $P \to N$. To see that 2. is equivalent to 3. note that free modules have property 3 and that property 3 is stable under retractions and for the converse note that every module is a quotient of a free module. –  t.b. Mar 15 '12 at 11:27
    
Thanks for 3. implies 1. But how is every module a quotient of a free module if you don't have a unit? –  Julian Kuelshammer Mar 15 '12 at 11:45
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Oh, you're right. Sorry in my work I almost always assume that modules are essential in the sense that $R \times M \to M$ is epi (since the rngs I consider have surjective multiplication $R \times R \to R$, this is a reasonable assumption for me to make). I was assuming this carelessly. I don't know about the more general situation. Sorry for the confusion. –  t.b. Mar 15 '12 at 12:00

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