Why $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module? 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?
 A: 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.
A: 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.
