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!


1st question answered in comments. For 2nd:

Lemma: Let $R$ be a domain s.t there is an $R$-module, $M$, which is nonzero, projective and injective $R$-module. then $R$ is a field.

$\mathbb{Q}$ 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.


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