Exercise 25, § 9

Let A be a ring and $\mathfrak{r}$ its radical. Prove that the following conditions are equivalent:

(i) Every left A-module admits a projective cover.

(i’) Every right A-module admits a projective cover.

(ii) The ring $A/\mathfrak{r}$ is semisimple, and for every sequence $(a_n)_{n\geqslant 1}$ of elements of $\mathfrak{r}$, there exists an integer $m$ such that $a_1\cdots a_m= 0$.

(Prove (ii)$\Rightarrow$(i) in the same way as the corresponding implication of Exercise 24, using Exercise 23. Assume that condition (i) is satisfied. Consider the submodule M of $A^{(\mathbf{N})}$ generated by $e_1-a_1e_2,e_2-a_2e_3, . .$.. Deduce from Exercise 16, e) applied to a projective cover of $A^{(\mathbf{N})}/M$ that we have $M = A^{(\mathbf{N})}$, which implies that $a_1\cdots a_n= 0$ for $n$ sufficiently large.)