Corollary

Let M be an A-module, $(x_i)_{i\in I}$ a generating family of M, and $x$ an element of M. The following properties are equivalent:

(i) We have $x\in \mathfrak{R}(M)$.

(ii) Every submodule N of M such that $N + Ax= M$ is equal to M.

(iii) For every family $(a_i)_{i\in I}$ of elements of A, the family $(x_i+a_ix)_{i\in I}$ generates the A-module M.

(i) $\Rightarrow$ (ii): Suppose that $x$ belongs to $\mathfrak{R}(M)$. Let N be a submodule of M such that $N + Ax= M$. We have $N +\mathfrak{R}(M) = M$; hence the A-module $M/N$ is equal to its radical (Corollary 1, b) of Proposition 1). Since it is monogenous, it is zero (Proposition 2), and we have M = N.

(ii) $\Rightarrow$ (iii): Let $(a_i)_{i\in I}$ be a family of elements of A. Denote by N the submodule of M generated by the family $(x_i+a_ix)_{i\in I}$. We have $x_i\in N + Ax$ for every $i\in I$, hence $N + Ax= M$. If property (ii) holds, then N is equal to M.

(iii) $\Rightarrow$ (i): Suppose that $x$ does not belong to $\mathfrak{R}(M)$. Then there exists a maximal submodule N of M that does not contain $x$. Since N is maximal, we have $N + Ax= M$; each of the elements $x_i$ can therefore be written as $y_i-a_ix$ with $y_i\in N$ and $a_i\in A$. The family $(x_i+a_ix)_{i\in I}$ is contained in N, hence does not generate M.