Proposition 2

Let M be a finitely generated A-module.

a) If M is not reduced to 0, then we have $\mathfrak{R}(M)\not= M$.

b) If N is a submodule of M such that $N +\mathfrak{R}(M) = M$, then we have N = M.

Let N be a proper submodule of M. By Proposition 3 of VIII, p. 49, there exists a maximal submodule L of M containing N. We have $N +\mathfrak{R}(M)\subset L$, and a fortiori $N +\mathfrak{R}(M)\not= M$. This proves b); the specific case N = 0 gives assertion a).