Proposition 3

Let A and B be rings and $f$ a homomorphism from A to B such that B is generated by the union of $f(A)$ and the commutant $f(A)'$ of $f(A)$ in B. Let M be a left B-module. Suppose that either the ring A is left Artinian or the A-module $f_*(M)$ deduced from M by restriction of scalars (II, §1, No. 13, p. 221) is finitely generated. We then have $\mathfrak{R}_A(f_*(M))\subset \mathfrak{R}_B(M)$.

Let S be a simple B-module. Suppose that either A is left Artinian or the A-module $f_*(S)$ is finitely generated. We will prove that $f_*(S)$ is without radical. For every $b\in f(A)'$, the homothety $b_S$ is an endomorphism of the A-module $f_*(S)$ and therefore preserves $\mathfrak{R}_A(f_*(S))$ (VIII, p. 152, Proposition 1). Since B is generated by $f(A)\cup f(A)'$, the radical $\mathfrak{R}_A(f_*(S))$ is a B-submodule of S, hence equal to 0 or S. If $f_*(S)$ is a finitely generated A-module, then we have $\mathfrak{R}_A(f_*(S))\not=f_*(S)$ (VIII, p. 153, Proposition 2). If the ring A is left Artinian, then we have $\mathfrak{R}_A(f_*(S)) =f(\mathfrak{r})S$, where $\mathfrak{r}$ is the radical of A (VIII, p. 174, Corollary of Proposition 2), and since $\mathfrak{r}$ is a nilpotent two-sided ideal of A (VIII, p. 173, Proposition 1), we cannot have $S =f(\mathfrak{r})S$. We therefore have $\mathfrak{R}_A(f_*(S)) = 0$ in both cases.

Let $u$ be a nonzero B-linear mapping from M to a simple M-module S. The mapping $u$ is surjective; consequently, if $f_*(M)$ is finitely generated, then so is $f_*(S)$. Under the assumptions of Proposition 3, the A-module $f_*(S)$ is without radical and $u$ is an A-linear mapping from $f_*(M)$ to $f_*(S)$. Hence, the kernel of $u$ contains $\mathfrak{R}_A(f_*(M))$ by Proposition 1 of VIII, p. 152. Since $u$ is arbitrary, we have $\mathfrak{R}_A(f_*(M))\subset \mathfrak{R}_B(M)$.