Let $f$ be a nonconstant polynomial with complex coefficients, and consider the ring $R = \Bbb C[x]/(f)$. I am trying to prove that $R$ has no nonzero nilpotent if and only if every finitely generated indecomposable $R$-module is projective module.
Any hints for this problem? I can't find a way to start the proof, in both directions
Note that $R$ has no non-zero nilpotents iff $f$ is square-free.
"$\Rightarrow$" In this case, by CRT we get that $R$ is a finite direct product of fields (that is, semisimple), and every $R$-module is projective.
"$\Leftarrow$" In particular, every simple $R$-module is projective. Then $R$ is semisimple (see here), that is, $R$ is a finite direct product of fields, and therefore it can't have non-zero nilpotent elements.
An elementary argument. First, let us determine the finitely generated indecomposable $R$-modules. Since $M$ is finitely generated over $R$ it is also a finitely generated $\mathbb C[X]$-module, and thus $$M\simeq\mathbb C[X]/(d_1)\oplus\cdots\oplus\mathbb C[X]/(d_r)$$ with $d_1\mid\cdots\mid d_r$. But $fM=0$, and hence $d_1\mid\cdots\mid d_r\mid f$. Since $M$ is indecomposable we have $r=1$.
Thus, we can write $M=\mathbb C[X]/(d)$ with $d\mid f$. Once again, since $M$ is indecomposable we have $d=p^m$ with $p$ irreducible and $m\ge 1$.
Now notice that $M$ can not be projective if $p^{m+1}\mid f$ (why?).
