# Free module implies projective module

Definiton : Let $M$ be an $A$-module. Then $M$ is projective if there exists an $A$-module $N$ such that $M \oplus N$ is free.

Prop: If $M$ is free then $M$ is projective.

Can we simply take the trivial module $\{0\}$ then $M \oplus \{0\} \cong M$ is free, so $M$ is projective?

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Yes, that argument works fine :)

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cheers! that kind of short arguments can make you doubt sometimes no? heh :) – user9467 Apr 12 '11 at 1:03
@Beginner: Yeah, I know what you mean – Zev Chonoles Apr 12 '11 at 1:07

The reason this seems simple is that there are many equivalent definitions of "projective module", and what you give as the definition is usually a property that is shown to be equivalent.

For example, in most treatments I know the the definition of projective module is given as either:

An $A$-module $P$ is projective if and only if for every $A$-modules $M$ and $N$, and every homomorphism $f\colon P\to N$, if $\varphi\colon M\to N$ is an onto module homomorphism, then there exists $g\colon P\to M$ such that $\varphi=f\circ g$.

or equivalently

An $A$ module $P$ is projective if and only if for every $A$-module $M$, $f\colon M\to P$ is an onto module homomorphism, then there exists a module homomorphism $g\colon P\to M$ such that $f\circ g=\mathrm{id}_P$; in particular, every module that has $P$ as a quotient can be written as a direct sum of $P$ and another module, and the direct sum is compatible with the original quotient map.

Under either of these definitions, the fact that free modules are projective is easy as well; one then proves that these two definitions are equivalent between them, and equivalent to the one you give.

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