(i) The A-module $A_s$ is isotypical.

(ii) The ring A is left Artinian, and every two-sided ideal of A is equal to 0 or A.

(iii) The ring A is left Artinian, and there exists a left A-module S that is simple and faithful.

If these conditions are satisfied, then the A-module $A_s$ has finite length and is isotypical of type S, and every simple A-module is isomorphic to S.

Let us prove that (i) implies (ii). Assume that (i) is satisfied. Then the finitely generated A-module $A_s$ is semisimple, hence has finite length and is Artinian (VIII, p. 71, Proposition 10); consequently, the ring A is left Artinian. The endomorphisms of the left A-module $A_s$ are the right multiplications by the elements of A. Since the left A-module $A_s$ is isotypical, it follows from Proposition 6, b) of VIII, p. 86 that the $(A$, A)-bimodule $_sA_d$ is simple. The sub-bimodules of $_sA_d$ are the two-sided ideals of A, so (i) implies (ii).

Let us prove that (ii) implies (iii). The ring A is not reduced to 0; consequently, there exists a simple A-module S. The annihilator of S is a proper two-sided ideal of A. If we assume that (ii) is satisfied, then that annihilator is equal to 0. The A-module S is then faithful, and (ii) implies (iii).

Let us prove that (iii) implies (i). Assume that (iii) is satisfied. Then there exists an integer $m\geqslant 1$ such that $A_s$ is isomorphic to a submodule of $S^m$ (VIII, p. 50, Proposition 5, a)). Since $S^m$ is an isotypical A-module of type S, the same holds for $A_s$ (VIII, p. 61, Proposition 2); hence, (iii) implies (i).

Suppose that conditions (i) through (iii) are satisfied. We have seen above that the A-module $A_s$ has finite length and is isotypical of type S. Every simple left A-module is isomorphic to a quotient of $A_s$, hence to S.

By Proposition 1, the A-module $A_s$ has finite length and is isotypical of type S. Hence, there exists an integer $m\geqslant 1$ such that the A-modules $A_s$ and $S^m$ are isomorphic. Then the A-module S is projective and finitely generated. It is generating (VIII, p. 80, Theorem 1), and Lemma 1 follows from Theorem 1 of VIII, p. 101, (ii)$\Rightarrow$(i) and (ii)$\Rightarrow$(iii) applied to the $(A,D)_{\mathbf{Z}}$-bimodule S.

We know that V is a simple E-module (VIII, p. 45, Example 3) and that its commutant is equal to $D_V$ (VIII, p. 82, Corollary 1). Let $(x_i)_{1\leqslant i\leqslant r}$ be a basis of V over the field D. The mapping $u\mapsto (u(x_i))_{1\leqslant i\leqslant r}$ is an isomorphism from the E-module $E_s$ to the E-module $V^r$; consequently, the E-module $E_s$ is isotypical of length $r$, so the ring E is simple.

Let us now prove Theorem 1. Recall (II, §10, No. 7, p. 349) that the ring $\mathbf{M}_r(D)$ can be identified with the endomorphism ring of the right D-vector space $D^r_d$; moreover, every right vector space of finite dimension $r$ over a field D is isomorphic to $D^r_d$ (II, §7, No. 1, p. 292). Theorem 1 therefore follows from Lemmas 1 and 2.

b) The opposite ring of a simple ring is simple.

c) The left length of a simple ring is equal to its right length.

Let D be a field, Z its center, and V a right vector space over D of finite dimension $r\geqslant 1$. We denote the endomorphism ring of V by E.

The mapping $z\mapsto z_V$ is an isomorphism from Z to the center of E by Corollary 2 of VIII, p. 83. Assertion a) follows. The dual $V^*$ of V is a right vector space over the opposite field $D^o$ of D, and its dimension is equal to $r$. The mapping $u\mapsto^tu$ is an isomorphism from the opposite ring $E^o$ of E to the ring End$_{D^o}(V^*)$. Consequently, the ring $E^o$ is simple, and the rings E and $E^o$ have the same left length, equal to $r$ (Lemma 2).

The condition is obviously sufficient.

Conversely, suppose that the rings $B =\mathbf{M}_r(D)$ and $B'=\mathbf{M}_{r'}(D')$ are isomorphic. Since $r$ is the length of $B_s$ and $r'$ that of $B'_s$ (Lemma 2), we have $r=r'$. Moreover, B is Morita equivalent to D and $B'$ to $D'$ (VIII, p. 102, Example 1). Consequently, the fields D and $D'$ are Morita equivalent, hence isomorphic (VIII, p. 111, Proposition 13, c)).

Let S be a simple left A-module; it is a finite-dimensional K-vector space over K. Its commutant is therefore an algebra of finite degree over K. The first assertion then follows from Lemma 1. If K is algebraically closed, then D = K by Theorem 1 of VIII, p. 47.

This follows from Lemma 1 of VIII, p. 120 and Morita’s theorem (VIII, p. 103).

a) Every A-module is projective and isotypical of type S, hence semisimple. If it has length $\mathfrak{a}$, then it is isomorphic to $S^{(\mathfrak{a})}$.

b) Every nonzero A-module is generating.

c) Two A-modules are isomorphic if and only if they have the same length.

Denote the opposite field of the commutant of S by D. By Lemma 1 of VIII, p. 120, S is an invertible $(A$, D)-bimodule. Let M be an A-module; by Lemma 3, it is isomorphic to a module of the form $S\otimes_DV$, where V is a left vector space over the field D.

The D-module V is projective and isotypical of type $D_s$; it is generating if and only if it is not reduced to 0. Finally, the length of $S\otimes_DV$ is equal to the dimension of the D-vector space V, and two vector spaces are isomorphic if and only if they have the same dimension. Proposition 2 now follows in view of Propositions 10 of VIII, p. 109 and 12 of VIII, p. 110.

Let $r\geqslant 1$ be an integer. We say that a cardinal $\mathfrak{a}$ is divisible by $r$ if there exists a cardinal $\mathfrak{b}$ such that $\mathfrak{a}=r\mathfrak{b}$. This is the case if $\mathfrak{a}$ is infinite because we have $r\mathfrak{a}=\mathfrak{a}($Set Theory, III, §6, No. 3, p. 188, Corollary 3). It follows from this remark that if the cardinal $\mathfrak{a}$ is divisible by $r$, then there exists a unique cardinal $\mathfrak{b}$ such that $\mathfrak{a}=r\mathfrak{b}$.

Every simple A-module is isomorphic to a quotient of $A_s$, hence finite-dimensional over $k$. The corollary then follows from Proposition 2, c).

(1) long$_A(M) =$ long(A) $\cdot$ dim$_A(M)$.

Suppose that M is free, and let $(e_i)_{i\in I}$ be a basis of M. The A-module M is the direct sum of the A-modules $Ae_i$, which are each isomorphic to $A_s$. Set $r=$ long$_A(A_s)$; this is an integer greater than or equal to 1 (VIII, p. 119, Proposition 1). We have long$_A(M) =r$ Card(I) by formula (13) of VIII, p. 73.

Conversely, suppose that the cardinal long$_A(M)$ is divisible by $r$. Let $\mathfrak{a}$ be the cardinal such that long$_A(M) =r\mathfrak{a}$. Then the A-module M has the same length as $A^{(\mathfrak{a})}_s$, hence is isomorphic to it by Proposition 2. This proves that M is free.

a) The mapping $a\mapsto a_M$ is an isomorphism from A to the endomorphism ring of the B-module M.

b) Suppose that M has finite length as an A-module. Then the ring B is simple, and we have

(2) long$_A(M) =$ long(B) and long$_B(M) =$ long(A).

The A-module M is generating by Proposition 2 of VIII, p. 122. By definition, we have $B = A'_M$, and assertion a) therefore follows from Theorem 2 of VIII, p. 82.

Suppose that M is an A-module of finite length. Choose a simple A-module S, and let D be the opposite field of the endomorphism ring of S. By Lemma 3, the A-module M is isomorphic to a module of the form $S\otimes_DV$, where V is a left vector space over the field D. The vector space V is finite-dimensional. By Theorem 3 of VIII, p. 64, the ring B is isomorphic to End$_D(V)$; by Lemma 2 (VIII, p. 121), the ring B is therefore simple. Taking Remark 1 of VIII, p. 63 into account, we obtain the equalities

long(B) = dim$_D(V) =$ long$_A(M)$.

By Remarks 1 of VIII, p. 63 and 3 of VIII, p. 121, we have the relations

long$_B(M) =$ long$_{End_D(V)}(S\otimes_DV) =$ dim$_D(S) =$ long(A),

which gives the last relation.

Let $r$ be the length of A; the A-module $A_s$ is isomorphic to $S^r$. Now, the left A-module B is isomorphic to $B\otimes_AA_s$ (II, §3, No. 4, p. 249), hence to $(B\otimes_AS)^r$ (II, §3, No. 7, p. 255, Proposition 7). We therefore have long$_A(B) =r$ long$_A(B\otimes_AS)$, and Proposition 5 follows from Proposition 3 of VIII, p. 123.

Let us introduce a basis $(e_i)_{i\in I}$ of C viewed as a left B-module and a basis $(f_j)_{j\in J}$ of B viewed as a left A-module. Then the family $(f_je_i)_{j\in J,i\in I}$ is a basis of C viewed as a left A-module (II, §1, No. 13, p. 222, Proposition 25); Proposition 6 follows.

b) The mapping $W\mapsto \mathfrak{b}(W)$ is a bijection from the set of linear subspaces of V to the set of right ideals of A.

c) Let $W_1$ and $W_2$ be linear subspaces of V. The relations $W_1\subset W_2$, $\mathfrak{a}(W_1)\supset \mathfrak{a}(W_2)$, and $\mathfrak{b}(W_1)\subset \mathfrak{b}(W_2)$ are equivalent.

Assertion b) follows from Example 1, b) of VIII, p. 104 applied to the invertible $(D^o,A^o$)-bimodule V, as does the equivalence of the relations $W_1\subset W_2$ and $\mathfrak{b}(W_1)\subset \mathfrak{b}(W_2)$.

Let $V^*$ be the dual of V, viewed as a right vector space over the opposite field $D^o$ of D. For any subspace W of V, denote the orthogonal of W in $V^*$ by $W'$. The mapping $W\mapsto W'$ is a bijection from the set of subspaces of V to the set of subspaces of $V^*$. If $W_1$ and $W_2$ are two subspaces of V, then the relations $W_1\subset W_2$ and $W'_1\supset W'_2$ are equivalent. Now, the mapping $u\mapsto^tu$ is an isomorphism from A to the opposite ring of End$_{D^o}(V^*)$; it transforms left ideals of A into right ideals of End$_{D^o}(V^*)$ and $\mathfrak{a}(W)$ into the set $\mathfrak{b}(W')$ of endomorphisms $h$ of $V^*$ such that $h(V^*)\subset W'$. Assertion a), as well as the equivalence of the relations $W_1\subset W_2$ and $\mathfrak{a}(W_1)\supset \mathfrak{a}(W_2)$, then follows from the assertion analogous to b) for the dual $V^*$ of V.

b) The minimal right ideals of A are the ideals $\mathfrak{b}(L)$, where L is a line in V. The maximal right ideals of A are the ideals $\mathfrak{b}(H)$, where H is a hyperplane in V.

Let $(L_i)_{i\in I}$ be a family of lines with direct sum V. Let $(\varepsilon_i)_{i\in I}$ be the family of projectors associated with the decomposition $V =\oplus_{i\in I}L_i$. The $\varepsilon_i$ are idempotent in A, and we have $\varepsilon_i\varepsilon_j= 0$ for $i\not=j$ and $\sum_{i\in I}\varepsilon_i= 1$. Denote the hyperplane $\sum_{j\not=i}L_j$ by $H_i$; it is the kernel of $\varepsilon_i$. We then have

The A-module $A_s$ is the direct sum of the family $(\mathfrak{a}(H_i))_{i\in I}$ of minimal left ideals, and $A_d$ is the direct sum of the family $(\mathfrak{b}(L_i))_{i\in I}$ of minimal right ideals.

Consider the specific case $V = D^n_d$, and identify A with the matrix ring $\mathbf{M}_n(D)$. Denote the interval $[1, n]$ in $\mathbf{N}$ by I and the canonical basis of V by $(v_i)_{i\in I}$; set $L_i=v_iD$, and denote by $E_{ij}$ the matrix units (II, §10, No. 3, p. 341). We then have $\varepsilon_i= E_{ii}$. The left ideal AE$_{ii}$ is equal to DE$_{1i}+\cdots +$ DE$_{ni}$ and consists of the matrices with all columns except the $i$-th equal to zero. The right ideal $E_{ii}A$ is equal to DE$_{i1}+\cdots +$ DE$_{in}$ and consists of the matrices with all lines except the $i$th equal to zero. We also have the relation

for $i$ and $j$ between 1 and $n$.