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I was reading about the torsion free abelian group. Does there exist a torsion free abelian group which is not projective but for which each of its torsion free homomorphic images are projective?

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Isn't the group itself a torsion-free homomorphic image? –  Martin Jan 29 '13 at 21:53
    
@Matin, sorry I did not mention in question but I want to exclude that case. –  Sharma Sharma Jan 29 '13 at 21:58
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I think $\mathbb{Q}$ might satisfy this, since the only torsion free homomorphic image it can have is itself or $0$, so if you're ruling out itself it's trivially true. –  John Stalfos Jan 31 '13 at 3:37

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John Stalfos's comment is correct: let's make it into an answer.

Consider $\mathbb{Q}$ as an abelian group (i.e., we work with the additive group). Then:

$\bullet$ $\mathbb{Q}$ is torsionfree.
$\bullet$ $\mathbb{Q}$ is divisible: for all $x \in \mathbb{Q}$ and all $n \in \mathbb{Z} \setminus \{0\}$, $x = ny$ for some $y \in \mathbb{Q}$ (indeed, for $y = \frac{x}{n}$).
$\bullet$ Any homomorphic image of a divisible abelian group is divisible.
$\bullet$ An abelian group is torsionfree and divisible iff it is (in a unique way) a $\mathbb{Q}$-vector space. Thus any torsionfree homomorphic image of $\mathbb{Q}$ must be a $\mathbb{Q}$-vector space $V$ and the map $q: \mathbb{Q} \rightarrow V$ is surjective and $\mathbb{Q}$-linear. By linear algebra, this can only happen if $V \cong \mathbb{Q}$ and $q$ is an isomorphism or if $V = 0$. So every torsionfree proper homomorphic image is projective.
$\bullet$ $\mathbb{Q}$ cannot be embedded into any free abelian group $\bigoplus_{i \in I} \mathbb{Z}$, since no nonzero element of a free abelian group is divisible by $n$ for all positive integers $n$. (The same reasoning shows that we cannot even embed $\mathbb{Q}$ in any direct product $\prod_{i \in I} \mathbb{Z}$. That is -- using terminology that I find rather unpleasant -- $\mathbb{Q}$ is a $\mathbb{Z}$-module which is torsionfree but not torsionless.) In particular $\mathbb{Q}$ is not projective.

(In fact projective modules over $\mathbb{Z}$ or any PID are necessarily free, but this fact is deeper than anything that took place above and was not used.)

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@ Pete thank you. This makes sense for me. –  Sharma Sharma Apr 3 '13 at 12:46

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