Let $R$ be a ring (not necessarily commutative, but if you have interesting things to say in the commutative case, please say them!) and $M$, $M''$ two finitely generated projective $R$-modules.

I was wondering if the kernel of a linear map $M \to M''$ is always a projective module.

Unless I'm mistaken, the question boils down to the a priori weaker one I've chosen as title. I guess the answer is no, but I'm unable to cook a counterexample (one first needs to take a linear form whose image isn't a projective submodule of $A$ but it's not enough).

Please make any comment you find relevant, for example is there are classes of rings over which this property is true (it's obviously so for hereditary rings and PIDs).


No, it isn't. If you take $R=K[x]/(x^2)$, then $R$ is projective over itself as always, but the kernel of the map $R\to R$ given by multiplication by $x$ is $Rx$ which isn't projective. To see this, note that $R$ is local, so the only projective modules are direct sums of copies of $R$, so the dimension (over $K$) of any projective module is divisible by $\dim{R}=2$, but $\dim{Rx}=1$.

  • $\begingroup$ Great answer. Do you have examples without zero divisors? $\endgroup$ – PseudoNeo Mar 22 '12 at 16:09
  • $\begingroup$ Hmm, interesting question. I can't think of anything off the top of my head. I think it reduces to the same question with projective replaced by free, so I imagine there are examples, but I don't know. $\endgroup$ – mdp Mar 22 '12 at 16:22

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