Suppose that M has finite length $d$. Then every strictly increasing or strictly decreasing sequence of submodules of M has at most $d+ 1$ terms (I, §4, No. 7, p. 44). Consequently, M is Artinian and Noetherian.

Conversely, suppose that M is Artinian and Noetherian. Let $\mathscr{S}$ be the set of submodules of M of finite length. The zero submodule is an element of $\mathscr{S}$, and since M is Noetherian, $\mathscr{S}$ has a maximal element N. Let us give a proof by contradiction and suppose that $M\not= N$. The set of submodules of M distinct from N and containing N then has a minimal element P because M is Artinian. The module $P/N$ has length 1, and since N is a module of finite length, the same holds for P (II, §1, No. 10, p. 212, Proposition 16). This contradicts the definition of N.

Begin by assuming that every submodule of M is finitely generated. Let $(P_n)_{n\in\mathbf{N}}$ be an increasing sequence of submodules of M, and let P be its union. This is a submodule of M. By assumption, there exists a finite subset F of M generating the module P; let $n\in \mathbf{N}$ be an integer such that $F\subset P_n$. We then have $P_n= P$, and the sequence $(P_n)_{n\in\mathbf{N}}$ is stationary. This proves that the module M is Noetherian.

The converse is a consequence of the following more precise statement.

Indeed, by VIII, p. 2, there exists a finite subset F of E such that $\sum_{x\in E}Ax=\sum_{x\in F}Ax$.

We will give the proof in the case of Artinian modules; the case of Noetherian modules is analogous.

Suppose that M is Artinian. Since every submodule of N is a submodule of M, the module N is Artinian. Let $(P_n)_{n\in\mathbf{N}}$ be a decreasing sequence of submodules of $M/N$. There exists a decreasing sequence $(Q_n)_{n\in\mathbf{N}}$ of submodules of M containing N such that $P_n= Q_n/N$ for every $n\in \mathbf{N}$ (I, §4, No. 6, p. 41, Theorem 4). Since M is Artinian, the sequence $(Q_n)$ is stationary, hence so is the sequence $(P_n)$. Consequently, the module $M/N$ is Artinian.

Conversely, suppose that the modules N and $M/N$ are Artinian, and consider a decreasing sequence $(P_n)$ of submodules of M. The sequence $P'_n= N\cap P_n$ of submodules of N is stationary. Likewise, the sequence $P''_n= (N + P_n)/N$ of submodules of $M/N$ is stationary. Hence, there exists an integer $m\in \mathbf{N}$ such that we have $P'_n= P'_m$ and $P''_n= P''_m$ for every integer $n\geqslant m$. The sequence $(P_n)$ is then stationary by the following lemma.

Let $x$ be an element of Q. It belongs to N + P; hence, there exists an element $y$ of P such that $x-y\in N$. Since Q contains P, the difference $x-y$ belongs to $N\cap Q$ and therefore to P. Consequently, $x$ belongs to P.

a) If the modules $M_i$ are Artinian (resp. Noetherian), then so is their sum $\sum_{i\in I}M_i$.

b) If the modules $M/M_i$ are Artinian (resp. Noetherian), then so is the module $M/\bigcap_{i\in I}M_i$.

By induction, it suffices to treat the case when $I =\{1,2\}$. The module $M_2/(M_1\cap M_2)$, quotient of $M_2$, is isomorphic to the submodule $(M_1+M_2)/M_1$ of $M/M_1$ (I, §4, No. 6, p. 41, Theorem 4).

In part a), we assume that $M_1$ and $(M_1+ M_2)/M_1$ are Artinian (resp. Noetherian); the same then holds for $M_1+ M_2$ (Proposition 3).

In part b), we assume that $M/M_2$ and $M_2/(M_1\cap M_2)$ are Artinian (resp. Noetherian); the same then holds for $M/(M_1\cap M_2)$ (loc. cit.).

We will use the following lemma in the proof.

We use induction on $n$. First, suppose $n= 1$. If M were not of finite length, then we could construct a sequence $(M_m)_{m\in\mathbf{N}}$ of submodules of M of length 1 with $M_m\not\subset (\sum_{i<m}M_i)$ for every $m\in \mathbf{N}$. We would then have $M_m\cap \sum_{i<m}M_i= 0$ for every $m$, and the sum of the family $(M_m)_{m\in\mathbf{N}}$ would be direct. However, this contradicts the fact that the A-module M is Artinian (VIII, p. 2, Example 2).

Now, suppose $n\geqslant 2$. Let $(M_i)_{i\in I}$ be a family of submodules of M of length $\leqslant n$ with sum M. For every $i\in I$, choose a submodule $M'_i$ of $M_i$ of length $\leqslant n-1$ such that $M_i/M'_i$ has length $\leqslant 1$. Set $M'=\sum M'_i$, and denote the image of $M_i$ in $M''= M/M'$ by $M''_i$. The modules $M''_i$ are of length $\leqslant 1$, and their sum is $M''$. The modules $M'$ and $M''$ are Artinian (VIII, p. 3, Proposition 3); by the induction hypothesis, they are of finite length. Hence, M has finite length (II, §1, No. 10, p. 212, Proposition 16).

Let us now prove Theorem 1. Let $\mathscr{S}$ denote the set of left ideals $\mathfrak{a}$ of A such that the module $A_s/\mathfrak{a}$ has finite length. Let $(\mathfrak{a}_i)_{i\in I}$ be a finite family of elements of $\mathscr{S}$. By Proposition 1 of VIII, p. 2, the A-module $A_s/\mathfrak{a}_i$ is Artinian and Noetherian for every $i\in I$. Consequently, $A_s/\bigcap_{i\in I}\mathfrak{a}_i$ is Artinian and Noetherian (VIII, p. 4, Corollary of Proposition 3), hence of finite length (VIII, p. 2, Proposition 1). This proves that $\mathscr{S}$ is a left directed set for the inclusion. Since the ring A is left Artinian, the set $\mathscr{S}$ has a least element $\mathfrak{b}$. We denote the length of the A-module $A_s/\mathfrak{b}$ by $n$.

Let $x$ be an element of $A_s$ and $\mathfrak{a}$ its annihilator (II, §1, No. 12, p. 219). The A-module $Ax$ is isomorphic to $A_s/\mathfrak{a}$. If $Ax$ has finite length, then $\mathfrak{a}$ belongs to $\mathscr{S}$, so $\mathfrak{a}$ contains $\mathfrak{b}$ and $Ax$ has length $\leqslant n$. Thus, every monogenous left ideal of A of finite length has length $\leqslant n$. Let $\mathfrak{c}$ be the sum of these ideals; this is a left ideal of A, of finite length by Lemma 3. Every left ideal of A of finite length is a sum of monogenous left ideals of finite length and is therefore contained in $\mathfrak{c}$. Hence, $\mathfrak{c}$ is the largest left ideal of A of finite length.

If $\mathfrak{c}$ were distinct from A, then the set of left ideals of A containing $\mathfrak{c}$ and distinct from $\mathfrak{c}$ would have a minimal element $\mathfrak{c}'$. The A-module $\mathfrak{c}'/\mathfrak{c}$ would then have length 1, and $\mathfrak{c}'$ would have finite length, which contradicts the fact that $\mathfrak{c}$ is maximal. We therefore have $\mathfrak{c}= A$; the A-module $A_s$ has finite length.

Let A be a left Artinian ring. By Theorem 1, the A-module $A_s$ has finite length. We then apply Proposition 1 of VIII, p. 2.

Let A be a left (resp. right) Artinian ring; the length of the A-module $A_s$ (resp. $A_d$) (I, §4, No. 7, p. 44) is called the left (resp. right) length of the ring A. When A is a commutative Artinian ring, these two lengths coincide and are simply called the length of A. When A is left and right Artinian but is not commutative, the left and right lengths of A are not necessarily equal (VIII, p. 14, Exercise 3).

b) Let A be a left Artinian ring and M a left A-module. The following properties are equivalent: the module M is finitely generated; the module M is Artinian; the module M has finite length; the module M is Noetherian.

Let us prove a). Every monogenous submodule of M is isomorphic to a quotient of $A_s$, hence is Noetherian by Proposition 3 of VIII, p. 3. The module M is a finite sum of such submodules; it is therefore Noetherian by the corollary (VIII, p. 4) of Proposition 3. Every submodule of M is then finitely generated (VIII, p. 3, Proposition 2)

Now, suppose that the ring A is left Artinian. We see, as in the previous section, that if the A-module M is finitely generated, then it is Artinian. If it is Artinian, then it has finite length: indeed, its monogenous submodules are isomorphic to quotients of $A_s$ and are therefore of finite length less than that of $A_s$, and the assertion follows from Lemma 3. Every module of finite length is Noetherian, and every Noetherian module is finitely generated. This proves b).

b) Let A be a left Artinian (resp. left Noetherian) ring, and let $\mathfrak{a}$ be a two-sided ideal of A; the ring $A/\mathfrak{a}$ is left Artinian (resp. left Noetherian).

c) Let $(A_i)_{i\in I}$ be a family of left Artinian (resp. left Noetherian) rings. The ring $\prod_{i\in I}A_i$ is left Artinian (resp. left Noetherian).

We will treat the case of Artinian rings; the case of Noetherian rings is analogous.

Let us prove a). By Proposition 4, the ring $B_s$ is an Artinian left A-module and a fortiori an Artinian left B-module.

Assertion b) follows from assertion a) applied to the canonical homomorphism from A to $A/\mathfrak{a}$.

Let us prove c). Set $A =\prod_{i\in I}A_i$. By assumption, $(A_i)_s$ is an Artinian left $A_i$-module and a fortiori an Artinian left A-module. By Example 5 of VIII, p. 4, the A-module $A_s$ is Artinian.

In any commutative ring, a maximal ideal is prime. Let A be an Artinian commutative ring. Let $\mathfrak{p}$ be a prime ideal of A. The ring $A/\mathfrak{p}$ is an integral domain and is Artinian (Proposition 5), hence is a field (VIII, p. 5, Example 3). Consequently, the ideal $\mathfrak{p}$ is maximal.

The polynomial ring $\mathbf{Q}[(X_n)_{n\in\mathbf{N}}]$ is an integral domain; it is not Noetherian (or Artinian) (VIII, p. 15, Exercise 9). It is a subring of its field of fractions, which is an Artinian (and Noetherian) ring.

Set E = End$_A(M)$. Let $(x_1, . . . , x_m)$ be a finite generating family of the E-module M. The mapping $\varphi :a\mapsto (ax_1, . . . , ax_m)$ from $(A_M)_s$ to $M^m$ is $A_M$-linear. Let $a$ be an element of $A_M$ such that $\varphi (a) = 0$. The set of elements $x$ of M such that $ax= 0$ is an E-submodule of M containing $x_1, . . . , x_m$ and is therefore equal to M, which implies $a= 0$.

This follows from Lemma 4 and Proposition 3 of VIII, p. 3.

a) Let M be a Noetherian A-module. The ring $A_M$ is Noetherian.

b) Let M be an A-module of finite length. The ring $A_M$ is Artinian.

Let M be an A-module. Under the assumptions of a) or b), the A-module M is finitely generated. Since A is commutative, $A_M$ is contained in the ring End$_A$(M), so that the countermodule of M is finitely generated. It then suffices to apply Proposition 6.

a) The addition in this ring is the usual addition of A[X].

b) The multiplication in this ring extends that of A.

c) The product in this ring of a sequence $(a,X, . . . ,X)$, consisting of an element $a$ of A followed by $n$ terms equal to X, is the polynomial $aX^n$.

d) In this ring, we have $Xa=\sigma (a)X +d(a)$ for every $a\in A$.

Let E be the endomorphism ring of the additive group A[X]. The mapping that sends $a\in A$ to the homothety $a_M$ of the left A-module M = A[X] is a ring homomorphism from A to E. Consider the elements $u,\sigma_M$, and $d_M$ of E defined by $u(\sum b_nX^n) =\sum b_nX^{n+1},\sigma_M(\sum b_nX^n) =\sum\sigma (b_n)X^n$, $d_M(\sum b_nX^n) =\sum d(b_n)X^n$. For every $a\in A$, we have

Set

It follows from (2) that for every $a\in A$, we have

Consider the mapping $\varphi : A[X]\rightarrow E$ defined by

This is a group homomorphism. An induction argument shows that we have $(X_M)^n(1) = X^n$ for every $n\in \mathbf{N}$. We therefore have $\varphi (P)(1) = P$ for every $P\in A[X]$, which proves that the homomorphism $\varphi$ is injective. We denote its image by B. The set B is a subgroup of E; it contains 1, and it is stable under left multiplication by $a_M$ for $a\in A$ and by $X_M$ (see (4)). It is therefore a subring of E. The unique ring structure on A[X] derived from that on B by transfer of structure via $\varphi$ has the properties of Proposition 7, where property d) results from relation (4).

If A[X] is endowed with a ring structure that has the properties of Proposition 7, then the left homothety $\boldsymbol{\gamma }_X$ of this ring (I, §8, No. 1, p. 97) necessarily sends $bX^n$ to $\sigma (b)X^{n+1}+d(b)X^n$ for $b\in A$ and $n\in \mathbf{N}$, hence is equal to $X_M$. The homothety $\boldsymbol{\gamma }_a$ is necessarily equal to $a_M$ for every $a\in A$. Consequently, we have $\boldsymbol{\gamma }_P=\varphi (P)$ for every $P\in A[X]$; the uniqueness in Proposition 7 follows.

The set A[X], endowed with the unique ring structure with the properties of Proposition 7, is denoted by $A[X]_{\sigma ,d}$ and called the polynomial ring in X with coefficients in A, relative to $\sigma$ and $d$. We simply denote it by $A[X]_{\sigma}$ when $d$ is the zero mapping and by A[X] when, moreover, $\sigma$ is the identity mapping on A. This notation is compatible with that introduced in IV, §1, No. 1, p. 1 for a commutative ring A.

Set $B = A[X]_{\sigma ,d}$. For any integer $n\geqslant 0$, we denote by $B_n$ the set of elements of B of the form $a_0+a_1X +\cdots +a_nX^n$. It is a left A-submodule of B. The mapping $\varphi_n: B_n\rightarrow A_s$ defined by $\varphi_n(a_0+a_1X +\cdots +a_nX^n) =a_n$ is A-linear.

Let $\mathfrak{b}$ be a left ideal of B. For every integer $n\geqslant 0$, the set $\mathfrak{a}_n=\varphi_n(\mathfrak{b}\cap B_n)$ is a left ideal of A. Since we have $Xa=\sigma (a)X +d(a)$ for every $a\in A$, we have

(6) $\varphi_{n+1}$(XQ) $=\sigma (\varphi_n(Q))$

for every $Q\in B_n$ and therefore $\sigma (\mathfrak{a}_n)\subset \mathfrak{a}_{n+1}$. Consequently, the sequence $\mathfrak{a}'_n=\sigma^{-n}(\mathfrak{a}_n)$ of ideals of A is increasing. Since the ring A is left Noetherian, there exists an integer $m\geqslant 0$ such that we have $\mathfrak{a}'_n=\mathfrak{a}'_{n+1}$ for $n\geqslant m$. Since $\sigma$ is surjective, we have the relation

for every integer $n\geqslant m$.

Let $\mathfrak{c}$ be the left ideal of B generated by $\mathfrak{b}\cap B_m$. Since the left A-module $B_m$ is finitely generated and the ring A is left Noetherian, the left A-module $\mathfrak{b}\cap B_m$ is finitely generated (VIII, p. 7, Proposition 4 a)). The left ideal $\mathfrak{c}$ is therefore generated by a finite subset of B. It is clear that it is contained in $\mathfrak{b}$. Let us prove that it is equal to $\mathfrak{b}$ by proving by induction that for every integer $n\geqslant 0$, we have

Relation (8) is true by construction for $n\leqslant m$. From now on, we assume that $n$ is an integer $\geqslant m$ such that $\mathfrak{b}\cap B_n\subset \mathfrak{c}$. Let P be an element of $\mathfrak{b}\cap B_{n+1}$. Then $\varphi_{n+1}(P)$ belongs to $\mathfrak{a}_{n+1}=\sigma (\mathfrak{a}_n)$, and there consequently exists an element Q of $\mathfrak{b}\cap B_n$ such that $\varphi_{n+1}(P) =\sigma (\varphi_n(Q))$. Set $R = P-$ XQ. Because of relation (6), we have $\varphi_{n+1}(R) = 0$, that is, $R\in B_n$. Since P and Q belong to the left ideal $\mathfrak{b}$ of B, the same is true for R; hence, R and Q belong to $\mathfrak{b}\cap B_n$, which is contained in the ideal $\mathfrak{c}$ by the induction hypothesis. Consequently, P belongs to $\mathfrak{c}$. This proves that we have $\mathfrak{b}\cap B_{n+1}\subset \mathfrak{c}$.

It follows that $\mathfrak{b}$ is equal to $\mathfrak{c}$; it is therefore a finitely generated ideal of B. This proves that the ring B is left Noetherian.

If the endomorphism $\sigma$ of the ring A is not an automorphism, then the ring $A[X]_{\sigma ,d}$ is not necessarily left Noetherian, even when A is a Noetherian commutative ring (VIII, p. 22, Exercise 26).

This follows by induction from Theorem 2, taking Proposition 8 of III, §2, No. 9, p. 453, into account.

Such an algebra is isomorphic to an algebra of the form $A[X_1, . . . ,X_n]/\mathfrak{a}$, where $n\geqslant 0$ and $\mathfrak{a}$ is an ideal of $A[X_1, . . . ,X_n]$. We then apply Corollary 1 and Proposition 5 of VIII, p. 7.

Indeed, let A be a commutative ring. The subrings of A generated (as $\mathbf{Z}$-algebras) by finitely many elements are Noetherian by Corollary 2. They form a right directed family of subrings of A, with union A.