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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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Didn't you use that $P$ is free over $\{u_{i}\}_{i\in I}$? –  t.b. Jul 23 '11 at 20:11
As the answers showed, i did, even though i was unaware of it:-) But still, i think the question was justified because i had in mind a non-free module exactly as i describe it and a map $\psi$ as i gave it. Now i see that what i was missing is that this $\psi$ is not necessarily a homomorphism, as the second answer shows. –  Manos Jul 23 '11 at 22:25
And i agree that i was not careful enough reading Lang's theorem, which assumes that the module is free since it mentions that $\left\{u_i\right\}_{i \in I}$ is in fact a basis. –  Manos Jul 23 '11 at 22:33
Manos, No need to explain: think of it this way: you'll never forget now that free modules are projective and you discovered it yourself :) –  t.b. Jul 23 '11 at 22:35
Thank you Theo, you are really encouraging :-) –  Manos Jul 23 '11 at 23:10

2 Answers 2

up vote 2 down vote accepted

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: as my answer shows, your suspicion is entirely correct. As you can see from the quoted statement of the theorem, he doesn't literally use the word "free" in the statement of the theorem. Rather here and in the rest of the discussion, the concept of "basis" is the privileged one. The existence of the current question suggests that this may be a slight expository error... –  Pete L. Clark Jul 23 '11 at 20:39
@Arturo: Thank you for your excellent answer; you exactly captured the point where i was confused. –  Manos Jul 23 '11 at 22:39

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. –  Manos Jul 23 '11 at 22:42

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