Proposition 6

For every A-module M, we have $\mathfrak{R}(A)M\subset \mathfrak{R}(M)$. We have equality if the A-module M is projective.

Let P be a maximal submodule of M; the A-module $M/P$ is simple, hence annihilated by $\mathfrak{R}(A)$ by Proposition 5 of VIII, p. 155. We therefore have $\mathfrak{R}(A)M\subset P$ for every maximal submodule P of M, hence $\mathfrak{R}(A)M\subset \mathfrak{R}(M)$.

We clearly have $\mathfrak{R}(A_s) =\mathfrak{R}(A)A_s$. If the A-module M is projective, then there exists an A-module N such that $M\oplus N$ is free, that is, the direct sum of a family $(L_i)_{i\in I}$ of modules isomorphic to $A_s$. By Corollary 2 of VIII, p. 152, we have $\mathfrak{R}(M\oplus N) =\mathfrak{R}(M)\oplus \mathfrak{R}(N)$ and $\mathfrak{R}(\bigoplus_{i\in I}L_i)=\bigoplus_{i\in I}\mathfrak{R}(L_i)$. We then deduce $\mathfrak{R}(M) =\mathfrak{R}(A)M$ from the equality $\mathfrak{R}(L_i) =\mathfrak{R}(A)L_i$.