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Let $\mathbb{F}$ be a field (of characteristic 0, if needed) and $\mathbf{V}$ an $\mathbb{F}$-vector space. Let $T\in\mbox{End}_\mathbb{F}(\mathbf{V})$ be an endomorphism, and let (following Bourbaki) $\mathbf{V}_T$ be the $\mathbb{F}[X]$-module given by $p(X)\cdot v=p(T)v$.

Question: Is there a non-trivial but tractable sufficient condition on $\mathbf{V}$ and/or $T$ for the torsion submodule of $\mathbf{V}_T$ to be a direct factor?

To be honest, I don't even know a counterexample where this fails to hold. For a general PID I found a counterexample here.

This is trivially true when $\dim\mathbf{V}<\infty$ because then $\mathbf{V}_T$ is torsion (Jordan normal form for matrices). By non-triviality I mean that $\mathbf{V}_T$ is not torsion.

Comments are welcome. I apologize for my illiteracy in algebra in advance.

Thank you.

Edit: Thanks for all comments. Now I have understood that one such condition is that $\mathbf{V}_T$ is finitely generated over $\mathbb{F}[X]$. What about the second, apparently more general case mentioned by Denis Nardin below? Is it true that if $\mathbf{V}_T/\mathbf{V}_T^{tors}$ is free then it is a direct factor?

Edit 2: Thanks again for the comments. It seems to me that every Noetherian torsion free module of a PID is necessarily free. Is there an example of a torsion free $\mathbb{F}[X]$-module that is not Noetherian?

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    $\begingroup$ I think it holds when $V_T$ is finitely generated as a $F[X]$-module. More generally, when $V_T/V^{tors}_T$ is a free $F[X]$-module $\endgroup$ Apr 7, 2016 at 23:24
  • $\begingroup$ When the dimension is finite, isn't the torsion submodule 0 (because torsion occurs when something is killed by a non-zero-divisor)? $\endgroup$ Apr 8, 2016 at 3:50
  • $\begingroup$ Thanks for the comments. I think when $\mathbf{V}_T$ is finitely generated then it is torsion, so it is the trivial case in my definition. Saying that $\mathbf{V}_T/\mathbf{V}_T^{tors}$ is a free module is too technical to be called a sufficient condition on $\mathbf{V}$ and $T$. One could say 'exaclty when the torsion submodule is a factor'. I am looking for something more naturally looking. Regarding the second comment, yes, torsion is when a non-zero has a kernel. This always happens in finite dimension, because $p(T)=0$ for $p$ the characteristic polynomial. $\endgroup$
    – Bedovlat
    Apr 8, 2016 at 10:01
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    $\begingroup$ Certainly the structure thm for f.g. modules over a PID ensures that if V is f.g. as a $F[X]$ module, then the torsion submodule is a direct factor. For an example of a f.g. module that is not torsion, you can just take $F[X]$ itself. That is, a countable-dimension vector space where $X$ acts by a "right shift". $\endgroup$ Apr 9, 2016 at 1:51
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    $\begingroup$ Certainly, if $\mathbf{V}_T/\mathbf{V}_T^\mathrm{tors}$ is free, it is projective, hence the exact sequence $0\to\mathbf{V}_T^\mathrm{tors}\to \mathbf{V}_T\to\mathbf{V}_T/\mathbf{V}_T^\mathrm{tors}\to 0$ splits. From this, you see that an apparently weaker condition is that the quotient be projective, but over a PID projective implies free (see wikipedia, projective module) hence no new modules occur. $\endgroup$ Apr 9, 2016 at 9:15

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