Corollary

Let A be a left Artinian ring. For every A-module M, the radical of M is equal to $\mathfrak{r}M$.

The radical $\mathfrak{R}(M)$ of M contains $\mathfrak{r}M$ (VIII, p. 158, Proposition 6). Moreover, the A-module $M/\mathfrak{r}M$ is annihilated by $\mathfrak{r}$. By Proposition 2, it is therefore without radical, which implies $\mathfrak{R}(M)\subset \mathfrak{r}M$ (VIII, p. 152, Corollary 1, c)).