2
$\begingroup$

This is an exercise from Elements of the representation theory of associative algebras, vol. 1 by Assem I., Simson D., Skowronski A.

Let $A=K[t]$ ($K$ is a field) and prove that the cyclic $A$-module $M=K[t]/(t^3)$ has no projective cover in $\text{Mod}A$.

Well, my idea was to suppose instead that there is some projective cover, so there exists a projective $A$-module $P$ and an epimorphism $h:P\rightarrow M$ such that if $g:N\rightarrow P$ is an $A$-homomorphism, $g$ is surjective if $gh$ is. Then I would find a strict submodule of $P$, perhaps using some ideal connected to $t^3$, to reach a contradiction. Perhaps I'm not making much sense, so if someone could help it would be appreciated.

$\endgroup$
5
$\begingroup$

Let us prove this in two steps:

  1. If $P\to M$ is a projective cover, then it is isomorphic (as a projective cover) to the natural projection $\pi:K[t]\to K[t]/(t^3)$;
  2. The natural projection $\pi$ is not a projective cover. We get a contradiction.

To prove 1., assume that $h:P \to M$ is a projective cover. Let $x\in P$ be a preimage of $1\in M$ by $h$, and let $f:K[t]\to P$ be the morphism defined by $f(1)=x$. Then the composition $fh$ is equal to $\pi$ and is thus surjective, and since $h$ is a projective cover, this implies that $f$ is surjective. The only (non-trivial) quotient of $K[t]$ which is projective is $K[t]$ itself; thus $f$ is an isomorphism, and the projective cover $h$ is isomorphic to $fh:K[t]\to K[t]/(t^3)$, which is equal to the natural projection $\pi$.

To prove 2., notice that the morphism $g:K[t]\to K[t]$ sending $1$ to $1+t^3$ is not surjective, but that $g\pi=\pi$ is. Thus $\pi$ cannot be a projective cover, and the proof is over.

Note that this proof works for any cyclic torsion module $M$ over a principal ideal domain $A$.

$\endgroup$

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.