In general, we have that:
free $\Rightarrow$ projective $\Rightarrow$ flat
injective $\Rightarrow$ divisible ( ($\Rightarrow$) be ($\Leftrightarrow$) in PIDs)
Simple Counter-examples:
projective but not free: $\mathbb{Z}_2$ is $\mathbb{Z}_6$ projective but not $\mathbb{Z}_6$ free
flat but not projective: $\mathbb{Q}$
My questions:
1) Please give counter-examples: divisible but not injective, flat but not injective.
2) In proof about $\mathbb{Q}$ is not a projective $\mathbb{Z}$-module, I use 2 properties:
First: $P$ is projective $\Leftrightarrow$ there is a free module $F$ and an $R$-module $K$ such that $F≅K⊕P$.
Second: Every submodule of a free module in PID is free.
The first is easy to prove but the second isn't. Other way to prove $\mathbb{Q}$ is not projective that use projective basis, but really it's difficult to understand for me.
So is there other way?
Thanks for regarding!
Best Answer
1st question answered in comments. For 2nd:
$\mathbb{Q}$ is a nonzero and injective $\mathbb{Z}$-module (as you mentioned injective $\Leftrightarrow$ divisible in PIDs). $\mathbb{Z}$ is a domain and not field. now use lemma.