TITLE: Why $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module?
QUESTION [7 upvotes]: 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?

REPLY [6 votes]: 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.