Divisible module proof: $\mbox{Ext}(R/aR; M) = 0, \forall a \in R \Rightarrow M $is divisible This is a problem from my lecture notes, and I don't really know how to tackle it. Any help will be appreciated.
The question reads as follow:
Problem
Given an integral domain $R$, and an $R$-module $M$. Is the following statement true:
If $\mbox{Ext}(R/aR; M) = 0, \forall a \in R$, then $M$ is divisible.

In fact, the problem has 2 parts, the first part asks to prove:
If $M$ is divisible, then $\mbox{Ext}(R/aR; M) = 0, \forall a \in R$, which I've finally managed to do it.
Back to the second part, I think the statement is true. And what I'm currently struggling is that I don't really know how to relate $M$, and $R$. I've tried is to find a special module $W_m$, for each $m \in M$, such that, when knowing this exact sequence splits $0 \rightarrow M \xrightarrow{\chi} W_m \xrightarrow{\sigma} R/aR \rightarrow 0$, I can prove that $m$ is divisible by $a$. But as I don't know how to relate $M$, and $R$, I don't think that's the correct way to do it.
Thanks you guys a lot,
And have a good day,
 A: Let $W_m$ be the quotient of $R\oplus M$ by the submodule generated by $(a,-m)$.  We have a map $\pi:W_m\rightarrow R/aR$ sending the class of $(x,y)$ to the class of $x$.  This map is well defined because if $(x,y)=(x',y')+r(a,-m)$, then $x-x'=ra\in aR$.  This is clearly surjective, and the kernel consists of the classes of $(x,y)$ such that $x\in aR$.  If $x=ar$, then $(x,y)$ lies in the same class as $(x,y)-r(a,-m)=(0,y+rm)$, so the kernel of $\pi$ consists of the classes of elements of the form $(0,y)$.
We also have a map $i:M\rightarrow W_m$ sending $y\in M$ to the class of $(0,y)$.  If $y\in \ker i$, then $(0,y)$ and $(0,0)$ are in the same class in $W_m$.  Hence there exists $r$ such that $(0,y)=r(a,-m)$, so $ra=0$.  Since $R$ is a domain and $a\neq 0$, it is clear that $r=0$. Consequently $y=0$.  Hence $i$ is injective.  By inspection the image of $i$ agrees with our description for the kernel of $\pi$.  Hence we have an extension $0\rightarrow M \rightarrow W_m \rightarrow R/aR \rightarrow 0$.  Since the Ext group vanishes, $W_m\cong M \oplus R/aR$; in particular we have a map $\phi:W_m\rightarrow M$ splitting the inclusion.
Note that $a(1,0)=(0,m)$ as elements of $W_m$.  Let $u=\phi(1,0)\in M$.  Then $au=\phi(0,m)=\phi(i(m))=m$.  Hence $m$ is divisible by $a$.
