# A module with no projective cover

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.

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 $A$.