I've been getting a little lost in algebra today. Let $M$ be a finitely generated $R[x]$-module where $R$ is a PID. There is a short exact sequence
$$0\to tM \to M \to F \to 0 $$
where $tM$ is the torsion submodule (consisting of those $m\in M$ with $p\cdot m=0$ for some non-zero $p\in R[x]$) and $F$ is torsion-free.
Question: Does the above sequence split? In other words, can I write $M\cong tM \oplus F$?
I believe the answer is yes if $R$ is a field, because then $R[x]$ is a PID and in that case finitely generated torsion-free modules should be projective. But what happens if $R$ is not a field? I don't expect a positive answer, in general, but I'm too dense to think of a counterexample.
By the way, I'm mostly interested in the case $R=\mathbb Z$, if it makes a difference. Any pointers are appreciated.
Best Answer
Let $r\in R$ be any nonzero nonunit element. Let $M$ be generated by two elements $a$ and $b$, with relations $r(ra+xb)=x(ra+xb)=0$. The torsion submodule $tM$ is generated by $ra+xb$ (this takes a little work to prove and uses the assumption that $r\neq 0$). Suppose $tM$ were a direct summand of $M$, so there is splitting map $f:M\to tM$. We would have $f(a)=p(ra+xb)$ and $f(b)=q(ra+xb)$ for some $p,q\in R[x]$, so $f(ra+xb)=(rp+xq)(ra+xb)=0$ since $r(ra+xb)=x(ra+xb)=0$. But we must have $f(ra+xb)=ra+xb$ since $ra+xb\in tM$, so this is a contradiction (here we use the assumption that $r$ is not a unit to be sure that $ra+xb$ is not $0$, since its annihilator $(r,x)$ is a proper ideal of $R[x]$).