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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.

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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 $R[x]$).

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  • $\begingroup$ Perfect example. Thanks! By the way, in order to prove that $ra+xb$ generates $tM$ I needed to assume that $r$ is prime. This is fine, because PIDs always have prime elements. But apparently, there are integral domains without prime elements. Did you have an argument in mind that works for general integral domains? $\endgroup$ Mar 25 '20 at 14:33
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    $\begingroup$ You can use the fact that $x$ is prime in $R[x]$ instead. Here's what I had in mind. It suffices to show that the quotient $M/(ra+xb)$ is torsion-free. Show that by showing it is isomorphic to the ideal $(r,x)\subset R[x]$ by mapping $a$ to $x$ and $b$ to $-r$. To show this map is injective, you have to show that if $xp-rq=0$ for $p,q\in R[x]$ then there exists $s$ such that $p=rs$ and $q=xs$. To find $s$, use primality of $x$ to conclude that $x$ divides $q$. $\endgroup$ Mar 25 '20 at 14:53
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    $\begingroup$ Another way to think of it is that $r$ and $x$ form a regular sequence in $R[x]$ and then what we are proving is basically the exactness of the associated Koszul complex. (This perspective is in fact how I came up with the example: the Koszul resolution of $(r,x)$ makes it clear that $\operatorname{Ext}^1((r,x),R/(r,x))$ is nontrivial, and then my module is just the explicit realization of such a nontrivial extension.) $\endgroup$ Mar 25 '20 at 14:57

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