Let $K$ be some field. I want to show that every indecomposable right $K[t]$-module $M$ is has a projective resolution of the form $$0 \to P_1 \to P_0 \to M \to 0.$$ To that end, I have a hint: Such a module $M$ must be of the form $$M = \displaystyle{\frac{K[t]}{\langle p(t) \rangle}}$$ with $\langle p(t) \rangle = p(t) K[t]$, for $p(t)$ the power of some irreducible polynomial in $K[t]$.
My attempt was to show that both $\langle p(t) \rangle$ and $K[t]$ are projective $K[t]$-modules. The latter is clear, as every ring is projective over itself, but the first is giving me trouble. So I have a couple of questions:
- Is this strategy the best one, or there is some other path I should take?
- Is there some hint on how to prove that $\langle p(t) \rangle$ is projective?
If I wanted to prove the hint (and I am clueless on how to prove it), could someone hint me on how to start it?