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From the fact that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=0$, how do we conclude that $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module?

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    What definitions of projective are you aware of? – Qiaochu Yuan Apr 26 '14 at 05:22
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    An $R$-module $P$ is said to be projective if for any exact sequence $A\rightarrow{B}\rightarrow{0}$ with $R$-mod. hom. $g:A\rightarrow{B}$ and $R$-mod. hom. $f:P\rightarrow{B}$ there exists $R$-mod. hom. $h:P\rightarrow{A}$ such that $gh=f$. –  Apr 26 '14 at 05:26
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    Take $A = \mathbb{Z}[x_{1}, \ldots, ]$, $B = \mathbb{Q}$. There is a surjection from $A$ to $B$ given by mapping $x_{i}$ to $\frac{1}{i}.$ So, we define $g$ to be this map, $f$ to be the identity. But there is no possible $h : \mathbb{Q} \rightarrow A$ because the only units are $\pm 1$. – Siddharth Venkatesh Apr 26 '14 at 06:01
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    Alternatively, you could check that a projective module must be a summand of a free module, and a free $\mathbb{Z}$-module is not divisible, and so cannot contain $\mathbb{Q}$. – Prahlad Vaidyanathan Apr 26 '14 at 06:22
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    http://math.stackexchange.com/questions/506256/prove-that-hom-bbbz-bbbq-bbbz-0-and-show-that-bbbq-is-not-a?rq=1 – WLOG Apr 26 '14 at 07:08

2 Answers2

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Alternatively, another fast way to prove by some Theorems is that “Any modules over a PID, it is a projective module if and only if it is a free module.”

So by contradiction, suppose Q is a projective Z-module. Because Z is a PID, Q is also a free Z-module But It's not.

Because for all submodules of Q \ {0}, they are not linearly independent over Z. And thus the only independent submodule of Q is {0}, which cannot span the whole Q. So Q cannot find a basis of Z-module. Then this will complete the proof.

We can simplify the linearly independence clarification above to show that

“For all q/p and n/m belongs to Q \ {0}, where p, m are not 0, q/p and n/m are not linearly independent over Z.”

Suppose a(q/p)+b(n/m)=0 for some a, b belong to Z. Take a=pn and b=-mq belong to Z. Then a(q/p)+b(n/m) =(pn)(q/p)+(-mq)(n/m) =qn - qn =0. Therefore, q/p and n/m are not linearly independent over Z.

In sum, because Z is a PID and Q is not a free Z-module, then Q is not a projective Z-module, too.

Damien John
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    Have you read the question? It asks to use a given property in order to prove that Q is not a projective Z-module. (-1) – user26857 Jan 08 '18 at 09:25
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I try to create a counterexample but later to find out there is some mistake in the condition I've made. Maybe someone can help to fix it or try to find another better counterexamples?? Anyway, here's my (false) counterexample I'd tried to find.

// Consider A→B→0 with Z-module the exact sequence with A=Z, B=Z, g:A→B is an epimorphism defined by g(z)=z+1 for all z belongs to Z.

Because f:Q→B can only be the zero map and h:Q→A as well. yet gh(q)=g(0)=1=/=0=f(q) for all q belongs to Q.

Thus there isn't exist any (actually it's because we've only have that one for us to check and however it doesn't fulfill.) homomorphism h:Q→A such that gh=f for this exact sequence to be commutative.

Therefore, Q is not an projective Z-module. //

The mistake is...even though g:A→B is onto in the example above, it does not form a module map...：( Maybe we need to try to find another g such that it is a Z-module epimorphism pair of hom. f and epi. g to make the counterexample exist.

Damien John
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