Decomposition of a module over an integral domain If $M$ is an $R$-module over an integral domain $R$, then must it be true that $M\cong \mathrm{Tor}(M) \oplus M/\mathrm{Tor}(M)$?  
I am interested in the case where $M$ has finite rank $n$, if that simplifies things. $M$ having rank $n$ means that $M$ has a maximal $R$-linearly independent set of size $n$.  
 A: This answer gives some explicit examples, illustrating the claims in the comment by egreg.
(1) First, a finitely generated example.
Let $R=k[x,y]$ for a field $k$. Let $N$ be the submodule of $R^2$ generated by $(y,-x)$, and let $N'$ be the submodule generated by $(xy,-x^2)$ and $(y^2,-xy)$.
Then $N$ is the kernel of the map $R^2\to R$ given by $(r,s)\mapsto rx+sy$, and so $R^2/N$ is torsion free.
Let $M=R^2/N'$. Then $M$ has a one-dimensional torsion submodule $N/N'$ 
generated by $(y,-x)$, with torsion free quotient $R^2/N$. But the inclusion $N/N'\to R^2/N'$ doesn't split , as all maps $R^2\to N/N'$ send $(y,-x)$ to zero.
(2) Second, an infinitely generated example over a PID (in fact, over $\mathbb{Z}$). I think the best known example is $\prod_p\mathbb{Z}/p\mathbb{Z}$, where the product is over all primes, but this doesn't have finite rank. 
There's a finite rank example that I found in Fuchs books on Infinite Abelian Groups:
Let $P=\prod_{n\geq1}\mathbb{Z}/4^n\mathbb{Z}$, the group of sequences whose $n$th term is from $\mathbb{Z}/4^n\mathbb{Z}$,  let $A$ be the subgroup 
$\bigoplus_{n\geq 1}\mathbb{Z}/4^n\mathbb{Z}$ of finite sequences, and let $G$ be the subgroup of $P$ generated by $A$ and the sequences
$$b_i=(0,\dots,0,1,2,2^2,2^3,\dots),$$
with the term $1$ in the $i$th position.
Then $A$ is clearly contained in the torsion subgroup of $G$, and the quotient $G/A$ is generated by the $b_i$, which (modulo $A$) have infinite order and satisfy the relations $2b_{i+1}=b_i$. So $G/A\cong\mathbb{Z}[\frac{1}{2}]$ is the torsion free quotient of $G$. 
However, the map $G\to G/A$ can't split, since $G$ has no elements divisible by $2^n$ for all $n$.
