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Referring to Lang's Algebra p.137 P1, an $A$-module $P$ is projective if the following property holds: given $A$-homomorphism f:P \longrightarrow M^{''} and surjective $A$-homomorphism g:M \longrightarrow M^{''}, then there exists $A$-homomorphism $h:P \longrightarrow M$ such that $f = g \circ h$.

I am confused as follows: let $P$ be any $A$-module (not necessarily projective) and let $f$, $g$ be as above. Now let $\left\{u_i\right\}_{i \in I}$ be a set of generators of $P$. For every $u_i$ there exists some (not necessarily unique) $\xi_i \in M$ such that $f(u_i)=g(\xi_i)$, since $g$ is surjective. Then according to the straightforward Theorem 4.1 p.135 (again from Lang) there is a unique $A$-homomorphism $\psi:P \longrightarrow M$ such that $\psi(u_i)=\xi_i$. But then this $\psi$ is such that $f = g \circ \psi$ and so by definition $P$ is projective. What am i missing?

Thank you :-)

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    Thank you Theo, you are really encouraging :-)2011-07-23

2 Answers 2

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You cannot guarantee that $\psi$ is a module homomorphism, because the relations that hold in $P$ among the elements $u_i$ need not hold in $M$.

For a simple example, take $A=\mathbb{Z}$, and your proposed module $P$ the (non-projective) module $\mathbb{Z}/2\mathbb{Z}$. Let M'' = P, $f$ the identity, and let $M=\mathbb{Z}$ with $g\colon \mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ the canonical projection.

Your proposal is: take a generating set for $P$; okay, I take $\{1+2\mathbb{Z}\}$. Then find some element in $M$ that maps to the image of $1+2\mathbb{Z}$; any odd $a\in\mathbb{Z}$ would work. Then you try to define $\psi$ by mapping $1+2\mathbb{Z}$ to $a$. But the only module homomorphism from $P$ to $\mathbb{Z}$ is the zero map, so no $\psi\colon P\to M$ maps $1+2\mathbb{Z}$ to a required pre-image. So this process does not work. The reason it does not work is that the potential preimages in $M$ don't satisfy the same relations that the $u_i$ do in $P$, in this case, that $1+2\mathbb{Z}$ satisfies $2(1+2\mathbb{Z}) = \mathbf{0}$.

I don't have Lang in front of me, but I suspect that Theorem 4.1 is for free modules (as that is the only situation in which the theorem holds in that generality).

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    @Arturo: Thank you for your excellent answer; you exactly captured the point where i was confused.2011-07-23
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Theorem III.4.1 on p. 135 of Lang's Algebra says:

Let $A$ be a ring and $M$ a module over $A$. Let $I$ be a nonempty set, and let $\{x_i\}_{i \in I}$ be a basis of $M$. Let $N$ be an $A$-module, and let $\{y_i\}_{i \in I}$ be a family of elements of $N$. Then there exists a unique homomorphism $f: M \rightarrow N$ such that $f(x_i) = y_i$ for all $i$.

I bolded the key word: basis. Earlier on that page Lang defines a basis, and says that a module which admits a basis is called free. A basis is a much stronger condition than a generating set! For any ring which is not a division ring, there will exist modules which do not have a basis, i.e., are not free.

The relation between this notion and the definition of projective goes in one direction: it shows that any free module is projective. The converse is, in general, far from being true. In fact determining when projective modules are free is quite a story: see e.g. $\S 3.5.4$ of my commutative algebra notes.

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    Thank you for your answer.2011-07-23