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I was working through a problem that was showing that for a field $F$, $F[x]$-modules correspond to pairs $(V,T)$ of vector spaces over $F$ and linear transformations on that vector space.

The last part of the question asks to find a finitely generated $F[x]$-module which is not projective.

The only examples I really know of modules which are not projective are ones like $\mathbb{Z}/2\mathbb{Z}$ as a $\mathbb{Z}$-module, where I understand it not to be projective because it doesn't have enough homomorphisms going out to $\mathbb{Z}$: every module homomorphism leaving $\mathbb{Z}/2\mathbb{Z}$ has to satisfy $$0=f(0)=f(2\cdot z)=2\cdot f(z)$$ and $2\cdot f(z)$ implies f(z)=0 in $\mathbb{Z}$. This is a problem since $\mathbb{Z}$ does project onto modules which $\mathbb{Z}/2\mathbb{Z}$ has nontrivial homomorphisms into (e.g.: itself).

I've been trying to think along similar lines for this case of finitely generated $F[x]$ modules, using the above correspondence with pairs $(V,T)$. I've been trying to think of a finite dimensional vector space $V$ and a linear transformation $T_V$ on it, and another vector space $W$ with transformation $T_W$ such that there aren't many $F[x]$-module homomorphisms $(V,T_V)\to (W,T_W)$.

I know such an $F[x]$-module homomorphism has to satisfy, among other things, that $f(T_V \cdot v)=T_W \cdot f(v)$. Using the correspondence with vector spaces and transformations, I've been trying to think of matrices $T_V$ and $T_W$ which only satisfy $f T_V = T_W f$ for very restricted choices of $f$, hoping for an analogous situation to the $\mathbb{Z}/2\mathbb{Z}$ example. I'm not so sure that's a good idea though since I don't know really understand how important it is to that example that there's a nontrivial element ($2$) that annihilates all of the elements.

So, does it make sense to be trying to think of examples this way for this case? Can this be done for arbitrary $F$ or should I be looking at a specific field to get an example? Does anyone have any examples? (I'd prefer to be able to generate my own, but since my current repository of non-projective modules is very sparse, I'd still appreciate any contributions.)

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up vote 3 down vote accepted

A finitely generated module over a PID (like $F[X]$) is projective if and only if its free. Certainly any free module is projective. On the other hand, if $M$ is your finitely generated module and $M$ is projective, then for every prime ideal $\mathfrak{p}$ of $F[X]$, $M_\mathfrak{p}$ (the localization of $M$ at $\mathfrak{p}$) is a finitely generated projective module over the local ring $F[X]_\mathfrak{p}$, hence is free. In particular, $M_\mathfrak{p}$ is torsion free for all such $\mathfrak{p}$. Since taking the torsion submodule is compatible with localization, this implies that $(M_{tor})_\mathfrak{p}=0$ for all such $\mathfrak{p}$, so $M_{tor}=0$. The structure theory for $M$ then implies that it is free. So all you need to do is find an $M$ that isn't free, and for that, just find one that isn't torsion free. For example, $F[X]/(f)$ where $f$ is any non-constant polynomial.

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Perhaps this is easier: if $F[X]/(f)$ is projective ($f$ non-constant) then the sequence $0\rightarrow(f)\rightarrow F[X]\rightarrow F[X]/(f)\rightarrow 0$ splits, which means that $F[X]/(f)$ injects into $F[X]$. But $F[X]$ is torsion free, so $F[X]/(f)$ (being non-zero and torsion) cannot inject into it. –  Keenan Kidwell May 10 '11 at 18:41
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Moreover, note that by the Cayley-Hamilton Theorem, given any linear transformation $T\colon V\to V$ that is not the zero linear transformation, there always exists a nonconstant $f\in F[X]$ such that $f(T)$ is the zero linear transformation on $V$, so the action of $F[X]$ on $(V,T)$ must factor through $F[X]/(f)$; so the projective modules are actually the difficult ones to find. –  Arturo Magidin May 10 '11 at 18:47
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