Exercise 21, § 8

Let $\mathfrak{r}$ be the radical of $A$. Now prove that the following conditions are equivalent:

$\alpha$) $A/\mathfrak{r}$ is semi-simple, and for every sequence $(a_n)_{n \geqslant 1}$ of elements of $\mathfrak{r}$, there exists an integer $n$ such that $a_1 \ldots a_n = 0$.

$\beta$) Every $A$-module admits a projective covering.

$\gamma$) For every $A$-module $M$, one has $\mathrm{dpl}_A(M) = \mathrm{dp}_A(M)$ (X, p. 202, exercise 7).

$\delta$) For every inductive system $(M_\alpha, f_{\beta\alpha})_{\alpha \in I}$ of $A$-modules, one has:

$\varepsilon$) Every flat $A$-module is projective.

$\varphi$) Every decreasing sequence of monogenic right ideals of $A$ is stationary.

(For the equivalence of $\alpha$) and $\beta$), cf. VIII, § 8, exercise 26. To prove that $\beta$) implies $\gamma$), note that $M$ admits a minimal projective resolution $P$ and that $P_n/\mathfrak{r}P_n$ is isomorphic to $\mathrm{Tor}_n^A(A/\mathfrak{r}, M)$. To prove $\varepsilon) \Rightarrow \varphi)$, let $(I_n)_{n \geqslant 1}$ be a decreasing sequence of monogenic right ideals, such that

In the free module $A^{(\mathbf{N})}$, consider the submodules $N_p$ generated by $e_1 - a_1 e_2, \ldots, e_p - a_p e_{p+1}$. Deduce from $\varepsilon$) that $N = \bigcup_p N_p$ is a direct factor in $A^{(\mathbf{N})}$; conclude that $I_{n+1} = I_n$ for $n$ sufficiently large.

For the implication $\varphi) \Rightarrow \alpha)$, use VIII, § 9, exercise 12.)