a) Let $\mathfrak{m}$ be a left ideal of A. The A-module $A_s/\mathfrak{m}$ is simple if and only if $\mathfrak{m}$ is a maximal left ideal.

b) Let M be a simple A-module, and let $x$ be a nonzero element of M. We have the equality $M = Ax$, the annihilator $\mathfrak{m}$ of $x$ is a maximal left ideal of A, and the mapping $a\mapsto ax$ defines, by passing to the quotient, an isomorphism from $A_s/\mathfrak{m}$ to M.

c) Let M be a nonzero A-module. If we have $M = Ax$ for every nonzero element $x$ of M, then M is simple.

The submodules of $A_s/\mathfrak{m}$ are of the form $\mathfrak{n}/\mathfrak{m}$, where $\mathfrak{n}$ is a left ideal of A containing $\mathfrak{m}$ (I, §4, No. 6, p. 41, Theorem 4); consequently, the A-module $A_s/\mathfrak{m}$ is simple if and only if we have $\mathfrak{m}\not= A$ and every left ideal $\mathfrak{n}$ of A containing $\mathfrak{m}$ satisfies $\mathfrak{n}/\mathfrak{m}= 0$ or $\mathfrak{n}/\mathfrak{m}= A_s/\mathfrak{m}$, that is, $\mathfrak{n}=\mathfrak{m}$ or $\mathfrak{n}= A$. This proves a).

Under the assumptions of b), $Ax$ is a nonzero submodule of M, hence is equal to M. Consequently, the mapping $a\mapsto ax$ defines, by passing to the quotient, an isomorphism from $A_s/\mathfrak{m}$ to M; the A-module $A_s/\mathfrak{m}$ is therefore simple, and the left ideal $\mathfrak{m}$ is maximal by a). This proves b).

Under the assumptions of c), let N be a nonzero submodule of M. If $x$ is a nonzero element of N, then we have $Ax\subset N$ and $Ax= M$, and therefore M = N. Hence M is simple.

Indeed, by Krull’s theorem (I, §8, No. 6, p. 104, Theorem 1), there exist maximal left ideals of A.

a) If M is simple, then $f$ is injective.

b) If N is simple, then $f$ is surjective.

c) If M and N are simple,then $f$ is an isomorphism.

The kernel of $f$ is a proper submodule of M, and the image of $f$ is a nonzero submodule of N.

a) If M is simple, then we have Ker($f$) $= 0$, so $f$ is injective.

b) If N is simple, then we have Im($f$) $= N$, so $f$ is surjective.

c) If M and N are simple, then $f$ is both injective and surjective.

If M is a simple A-module, then every nonzero element of the nonzero ring End$_A(M)$ is invertible (Proposition 2, c)), so End$_A(M)$ is a field.

Let E be the endomorphism ring of the A-module M; it is a field by the corollary of Proposition 2 and an algebra over the field K. If we view M as a left vector space over the field E, then we have dim$_KM =$ (dim$_EM$)$[E : K]$ by Proposition 25 of II, §1, No. 13, p. 222, hence dim$_KM\geqslant [E : K]$. Since dim$_KM<$ Card(K) by assumption, the equality $E = K\cdot 1_M$ is then a consequence of the following lemma.

Let $x$ be an element of E - K and L the (commutative) subfield of E generated by $K\cup \{x\}$. Since K is algebraically closed, $x$ is transcendent over K. By VII, §2, No. 3, p. 10, Theorem 2 and p. 11, the elements $(x-\alpha )^{-1}$ of L, where $\alpha$ runs through K, are linearly independent over K. We therefore have $[E : K]\geqslant [L : K]\geqslant$ Card(K).

Since the ring A is commutative, $a_M$ is an endomorphism of the A-module M for every $a\in A$. By Theorem 1, we therefore have $A_M= K\cdot 1_M$, and M is a simple K-module, that is, a 1-dimensional vector space over the field K.

Let I be a set of cardinal strictly greater than the dimension of M over K (for example, the set of subsets of M). Let L be an algebraic closure of the field $K((X_i)_{i\in I})$ (V, §4, No. 3, p. 23, Theorem 2). Choose a K-linear form $\varphi$ on L such that $\varphi (1) = 1$, and let $v: M_{(L)}\rightarrow M$ be the K-linear mapping characterized by $v(\alpha \otimes m) =\varphi (\alpha )m$. Let $u$ be an endomorphism of M. The dimension of $M_{(L)}$ over L is equal to that of M over K and is strictly less than the cardinal of L. By Theorem 1, the endomorphism $1_L\otimes u$ of the $A_{(L)}$-module $M_{(L)}$ is of the form $\lambda \otimes 1_M$ with $\lambda \in L$. For every $x\in M$, we have

so that $u$ is the homothety $\varphi (\lambda )_M$.

Let N be a proper submodule of M. Denote by $\mathscr{S}$ the set of proper submodules of M containing N, ordered by inclusion; let us prove that $\mathscr{S}$ is inductive. Let $\mathscr{F}$ be a totally ordered subset of $\mathscr{S}$. If $\mathscr{F}$ is empty, then N is an upper bound for $\mathscr{F}$ in $\mathscr{S}$. In the opposite case, denote by Q the union of the elements of $\mathscr{F}$. Then Q is a submodule of M. Let F be a finite generating subset of M. If Q were equal to M, then F would be contained in a submodule $P\in \mathscr{F}$, which would imply P = M, contrary to the definition of $\mathscr{F}$. We therefore have $Q\in \mathscr{S}$, which proves that $\mathscr{S}$ is inductive. Proposition 3 then follows from Corollary 1 of Set Theory, III, §2, No. 4, p. 155.

Let N be a maximal submodule of M (Proposition 3), and let $\mathfrak{a}$ be the annihilator of the simple A-module $M/N$; it is a two-sided ideal of A, and we have $\mathfrak{a}(M/N) = 0$, whence $\mathfrak{a}M\subset N$ and consequently $\mathfrak{a}M\not= M$.

Since the ring A is commutative, the annihilator of a simple A-module is a maximal ideal of A (VIII, p. 46, Proposition 1). By Corollary 1 applied to the A-module M, there exists a maximal ideal $\mathfrak{a}$ of A such that $\mathfrak{a}M\not= M$. But $\mathfrak{a}M$ is a submodule of the simple B-module M; we therefore have $\mathfrak{a}M = 0$ and consequently $\mathfrak{a}\subset \mathfrak{m}$. Since the ideal $\mathfrak{m}$ is distinct from A, it is equal to $\mathfrak{a}$, so that $\mathfrak{m}$ is maximal. The module M is a finitely generated module over the field $A/\mathfrak{m}$; the last assertion follows.

Let M be a faithful simple A-module. Let $\alpha$ be a nonzero element of $\mathfrak{a}$. Since the A-module M is faithful, there exists an element $x$ of M such that $\alpha x\not= 0$. The homomorphism $a\mapsto ax$ from $\mathfrak{a}$ to M is then nonzero; it is therefore an isomorphism by Proposition 2 of VIII, p. 47.

a) There exist a natural number $m$ and an isomorphism from $A_s$ to a submodule of $M^m$.

b) If M admits a Jordan–Hölder series $(M_i)_{0\leqslant i\leqslant n}$, then every simple A-module is isomorphic to one of the quotients $M_i/M_{i+1}$.

By VIII, p. 2 applied to the A-module $A_s$ and the annihilators of the elements of M, there exist elements $x_1, . . . , x_m$ of M such that the annihilator of M is the intersection of the annihilators of the $x_i$. Since the A-module M is faithful, the A-linear mapping $a\mapsto (ax_1, . . . , ax_m)$ from $A_s$ to $M^m$ is injective, whence a).

Under the assumption of b), every simple quotient of $A_s$ is isomorphic to a quotient of a Jordan–Hölder series of $M^m$ (I, §4, No. 7, p. 43, Corollary), hence to one of the modules $M_i/M_{i+1}$ (I, §4, No. 7, p. 43, Theorem 6). We conclude that every simple A-module is isomorphic to a quotient of $A_s$.