Projective resolution of $R/rR$ R ring, and $R/rR$ as a $R$-module, what is the best projective module projecting to $R/rR$ to get a projective resolution of $R/rR$?
Is $R$ a free $R$-module? If it is, how should I view it as free?
 A: I will assume that $R$ is a ring with unit and that homorphisms preserve the unit. Let us try and prove that $R$ is free $R$-module on one generator. In particular let us show that the map $i:\{*\} \to R$ sending $*$ to $1$ has the following universal property: For each $R$-module $M$ and each map $f:\{*\}\to M$ there exists a unique homomorphism $\bar f: R \to M$ such that $\bar{f} i =f$. Suppose $M$ is an $R$-module and $f:\{*\} \to M$ is a map. If $\bar f$ exists, satisfies $\bar fi=f$, and is a homomorphism, then we must have $\bar{f}(1)=\bar{f}i(*)=f(*)$ and hence $\bar{f}(r)=\bar{f}(r1)=r\bar{f}=rf(*)$. This proves that if such a $\bar f$ exists it is unique a must be defined by the previous formula. Therefore we need only check that if $\bar f$ is defined by $\bar{f}(r) = rf(*)$, then it is a homorphism. We have $\bar{f}(r+r')=(r+r')f(*)=rf(*)+r'f(*)=\bar{f}(r)+\bar{f}(r')$ and $\bar{f}(rr')=(rr')f(*)=r(r'(f*))=r\bar{f}(r')$ as required. It follows that $R$ is a projective and that the canonical quotient $q: R \to R/rR$ is a surjective homorphism with domain projective.
