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.)


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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Keenan Kidwell May 10 '11 at 18:41
  • 2
    $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 10 '11 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.