We know (VIII, p. 156, Theorem 1) that every nilpotent two-sided ideal of A is contained in $\mathfrak{r}$. Let us prove that $\mathfrak{r}$ is nilpotent. Since the ring A is left Artinian and the sequence of two-sided ideals $\mathfrak{r}^n$ is decreasing, there exists an integer $p\geqslant 0$ such that we have $\mathfrak{r}^p=\mathfrak{r}^{p+1}$. Since A is Artinian, the A-module $A_s$ has finite length (VIII, p. 6, Theorem 1) and is therefore Noetherian. The left ideal $\mathfrak{r}^p$ is therefore finitely generated. By Nakayama’s lemma (Theorem 2 of VIII, p. 158), it follows that $\mathfrak{r}^p= 0$.

The ring $A/\mathfrak{r}$ is without radical (VIII, p. 155, Proposition 5); since it is left Artinian, it is semisimple (VIII, p. 154, Proposition 4).

A semisimple ring is left Artinian and without radical (VIII, p. 154, Proposition 4); it therefore has no nilpotent two-sided ideals other than 0. The converse follows from Proposition 1.

If A is left Artinian, then the radical is a nilpotent two-sided ideal (Proposition 1). Every left nil ideal is contained in the radical by Jacobson’s theorem (VIII, p. 156, Theorem 1). Corollary 2 follows.

(i) M is semisimple.

(ii) M is without radical.

(iii) M is annihilated by the radical $\mathfrak{r}$ of A.

We know that (i) implies (ii) (VIII, p. 153, Proposition 3). On the other hand, $\mathfrak{r}M$ is contained in the radical of M (VIII, p. 158, Proposition 6), so (ii) implies (iii).

Suppose that the A-module M is annihilated by $\mathfrak{r}$. We can view it as a module over the ring $A/\mathfrak{r}$. Now, the ring $A/\mathfrak{r}$ is semisimple (VIII, p. 173, Proposition 1), and every module over a semisimple ring is semisimple (VIII, p. 138, Proposition 4). Consequently, M is a semisimple module over the ring $A/\mathfrak{r}$ and a fortiori over the ring A. Hence (iii) implies (i).

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)).

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 $\mathfrak{R}_A(f_*(M))\subset \mathfrak{R}_B(M)$.

By Proposition 3, the radical $\mathfrak{R}_A(B_s)$ is contained in $\mathfrak{R}_B(B_s)$, which is nothing but the radical of the ring B. Moreover, we have $\mathfrak{R}(A)B\subset \mathfrak{R}_A(B_s)$ by Proposition 6 of VIII, p. 158. The corollary follows.

Let A be a left Artinian ring, and let M be a module over A. The radical $\mathfrak{r}$ of A is a nilpotent two-sided ideal, and the ring $A/\mathfrak{r}$ is semisimple (VIII, p. 173, Proposition 1). It follows that the $A/\mathfrak{r}$-module $M/\mathfrak{r}M$ is projective. The assertion then follows from Proposition 11 of VIII, p. 164.

a) Let P be a projective A-module. Denote the canonical mapping from P to $P/\mathfrak{r}P$ by $u$. Then $(P, u)$ is a projective cover of $P/\mathfrak{r}P$. In particular, the A-module P is finitely generated if and only if $P/\mathfrak{r}P$ is.

b) Let M be a module over the ring $A/\mathfrak{r}$. Then M, viewed as an A-module, has a projective cover. If $(P, u)$ is such a projective cover, then when passing to the quotient, $u$ induces an isomorphism from $P/\mathfrak{r}P$ to M. Moreover, P is indecomposable if and only if M is simple.

c) Let M and $M'$ be modules over the ring $A/\mathfrak{r}$. Let $(P, u)$ and $(P', u')$ be projective covers of the A-modules M and $M'$. Then M and $M'$ are isomorphic if and only if P and $P'$ are.

The radical $\mathfrak{r}$ of A is a nilpotent two-sided ideal (VIII, p. 173, Proposition 1). Assertion a) therefore follows from Corollary 1 of VIII, p. 163 and Remark 2 of VIII, p. 161.

Let us prove assertion b). The existence of a projective cover $(P, u)$ of M follows from Proposition 4 and the isomorphism from $P/\mathfrak{r}P$ to M of Proposition 9 of VIII, p. 162. Since the ring $A/\mathfrak{r}$ is semisimple, a module over this ring is indecomposable if and only if it is simple. The last assertion then follows from Corollary 2 of VIII, p. 160.

Let us prove c). Let $f$ be an isomorphism from P to $P'$; it induces an isomorphism $\overline{f}$ from $P/\mathfrak{r}P$ to $P'/\mathfrak{r}P'$. By b), these quotients are isomorphic to M and $M'$, respectively. Therefore, M is isomorphic to $M'$. Conversely, by Proposition 8 of VIII, p. 161, every isomorphism $f$ from M to $M'$ lifts to an isomorphism $\widetilde{f}$ from P to $P'$ such that $f\circ u=u'\circ \widetilde{f}$.

Let A be a left Artinian ring and $\mathfrak{r}$ its radical. Denote the set of classes of simple A-modules by $\mathscr{S}$; it is finite (VIII, p. 51). By Proposition 2 of VIII, p. 174, it can be canonically identified with the set of classes of simple modules over the semisimple ring $A/\mathfrak{r}$. For every $\lambda \in \mathscr{S}$, let $S_{\lambda}$ be a simple A-module of class $\lambda$, and choose a projective cover $(P_{\lambda}, u_{\lambda})$ of $S_{\lambda}$ (VIII, p. 175, Proposition 4). By Proposition 5, the A-module $P_{\lambda}$ is projective and indecomposable, and $u_{\lambda}$ defines an isomorphism from $P_{\lambda}/\mathfrak{r}P_{\lambda}$ to $S_{\lambda}$. Moreover, if P is a projective and indecomposable A-module, then there exists a unique $\lambda \in \mathscr{S}$ such that P is isomorphic to $P_{\lambda}:$ it is the unique $\lambda \in \mathscr{S}$ such that $P/\mathfrak{r}P$ is isomorphic to $S_{\lambda}$.

a) The A-module $\overline{P}= P/\mathfrak{r}P$ is semisimple, and the A-module P is isomorphic to $\bigoplus_{\lambda\in\mathscr{S}}P^{([P:\lambda])}_{\lambda}$.

b) Let $(Q_i)_{i\in I}$ be a family of indecomposable projective submodules of P with direct sum P. Then for every $\lambda \in \mathscr{S}$, the cardinal of the set $I(\lambda )$ of $i\in I$ such that $Q_i$ is isomorphic to $P_{\lambda}$ is equal to $[P :\lambda ]$.

The fact that $\overline{P}$ is semisimple follows from Proposition 2 (VIII, p. 174). The A-module $Q =\oplus_{\lambda\in\mathscr{S}}P^{([P:\lambda])}_{\lambda}$ is projective. Since $P_{\lambda}/\mathfrak{r}P_{\lambda}$ is isomorphic to $S_{\lambda}$, the quotient $Q/\mathfrak{r}Q$ is isomorphic to $\oplus_{\lambda\in\mathscr{S}}S^{([P:\lambda])}_{\lambda}$, that is, to $\overline{P}= P/\mathfrak{r}P$. By Proposition 5, the A-modules P and Q are isomorphic.

Suppose given a family $(Q_i)_{i\in I}$ of projective and indecomposable submodules with direct sum P. By Proposition 5, the A-module $Q_i/\mathfrak{r}Q_i$ is simple for every $i\in I$. For $\lambda \in \mathscr{S}$, denote by $I(\lambda )$ the set of $i\in I$ such that $Q_i/\mathfrak{r}Q_i$ is isomorphic to $S_{\lambda}$, hence to $P_{\lambda}/\mathfrak{r}P_{\lambda}$; it is also the set of $i\in I$ such that $Q_i$ is isomorphic to $P_{\lambda}$. Since $\overline{P}= P/\mathfrak{r}P$ is the direct sum of the family $(Q_i/\mathfrak{r}Q_i)_{i\in I}$, we have Card(I($\lambda$ )) $= [P :\lambda ]$ by Theorem 1 of VIII, p. 32.