Exercise 1, § 1

Tag 000X, permanent. Cite it as [Bourbaki, Tag 000X].

Algebra, VIII, p. 14

Let A be a principal ideal domain that is not a field and P a system of representatives consisting of irreducible elements of A. An A-module M is Artinian if and only if it is a torsion module and there exists a finite subset S of P such that the following two conditions are satisfied:

(i) For every $\pi \in P$ - S, the $\pi$ -primary component $M_{\pi}$ of M (VII, §2, No. 2, p. 8) is reduced to 0.

(ii) For every $\pi \in S$ , the set of elements of M annihilated by $\pi$ , viewed as a vector space over the field $A/\pi A$ , is finite-dimensional.

Solution

This solution was written by gpt-5-6-mini, gpt-5-6 and judged by a machine. It is not Bourbaki's and it has not been checked by a person. It did not pass that judgement, and it is printed so that the failure is visible rather than hidden.

Suppose first that $M$ is Artinian. Since $A$ is an integral domain which is not a field, and $M$ is Artinian, we show first that $M$ is torsion. If $x\in M$ is not torsion, then the homomorphism $A\to Ax$ , $a\mapsto ax$ , is injective, so $Ax\simeq A$ . By Proposition 3 of §1, $Ax$ is Artinian, and hence $A$ would be Artinian. This contradicts Example 3 of §2. Thus $M$ is a torsion module.

By the standard primary decomposition theorem for torsion modules over a principal ideal domain of VII, §2, No. 2, p. 8, we have

M=\bigoplus_{\pi\in P}M_\pi .

M_\pi\neq0

for infinitely many

\pi\in P

, these nonzero submodules have direct sum in

M

, contrary to Example 2 of §1. Hence there is a finite subset

S\subset P

such that

M_\pi=0

for

\pi\in P-S

. This proves (i).

Let $\pi\in S$ . By Proposition 3 of §1, the submodule $M_\pi$ is Artinian. The submodule

M_\pi[\pi]=\{x\in M_\pi\mid \pi x=0\}

is therefore Artinian. It is a vector space over the field

A/\pi A

, and hence it is finite-dimensional: indeed, an infinite-dimensional vector space is not Artinian by Example 2 of §1. Moreover,

M[\pi]=M_\pi[\pi].

Indeed, if

x=\sum_{\rho}x_\rho\in M

is annihilated by

\pi

, with

x_\rho\in M_\rho

, then for

\rho\ne\pi

some power

\rho^n

annihilates

x_\rho

, while

\pi

also annihilates

x_\rho

. Since

\pi

and

\rho^n

are relatively prime, there exist

u,v\in A

such that

u\pi+v\rho^n=1

, and therefore

x_\rho=0

. Thus

x=x_\pi\in M_\pi[\pi]

, while the reverse inclusion is immediate. Condition (ii) follows.

Conversely, suppose that $M$ is a torsion module and that there is a finite subset $S\subset P$ satisfying (i) and (ii). By the standard primary decomposition theorem for torsion modules over a principal ideal domain of VII, §2, No. 2, p. 8, we have

M=\bigoplus_{\pi\in S}M_\pi .

For each

\pi\in S

, condition (ii), together with the equality

M[\pi]=M_\pi[\pi]

proved above, says that

M_\pi[\pi]

is finite-dimensional over

A/\pi A

Fix $\pi\in S$ . We prove that $M_\pi$ is Artinian. Put $K=\operatorname{Frac}(A)$ and let

E_\pi=\{y\in K/A\mid \pi^ny=0\text{ for some }n\geqslant0\}.

Thus

E_\pi

is the

\pi

-primary component of

K/A

. It is an injective

A

-module: indeed, it is divisible, since division by every nonzero element of

A

is possible in

K/A

within its

\pi

-primary component, and a divisible module over a principal ideal domain is injective (equivalently, this follows directly from Baer's criterion, since every ideal of

A

is principal).

Moreover, $E_\pi[\pi]$ is one-dimensional over $A/\pi A$ , and $E_\pi$ is Artinian. To see the latter directly, every nonzero element of $E_\pi$ has the form $u/\pi^n+A$ , with $u$ relatively prime to $\pi$ ; it generates $\pi^{-n}A/A$ . Consequently every submodule of $E_\pi$ is one of

0,\quad \pi^{-1}A/A,\quad \pi^{-2}A/A,\quad\ldots,\quad E_\pi,

and these submodules form a descending chain. Thus

E_\pi

is Artinian.

Let $r=\dim_{A/\pi A}M_\pi[\pi]$ , and choose a basis $e_1,\ldots,e_r$ of $M_\pi[\pi]$ . Since $E_\pi[\pi]$ is one-dimensional, there is an injective homomorphism

M_\pi[\pi]\longrightarrow E_\pi^r

whose image is

(E_\pi[\pi])^r

. Since

E_\pi^r

is injective, this homomorphism extends to a homomorphism

f:M_\pi\longrightarrow E_\pi^r.

The submodule

M_\pi[\pi]

is essential in

M_\pi

: if

0\ne x\in M_\pi

, choose

n

minimal with

\pi^nx=0

; then

\pi^{n-1}x

is a nonzero element of

M_\pi[\pi]

. Hence

\ker f=0

, because

f

is injective on

M_\pi[\pi]

. Thus

M_\pi

is isomorphic to a submodule of

E_\pi^r

. Since

E_\pi^r

is Artinian by Example 5 of §1,

M_\pi

is Artinian by Proposition 3 of §1.

Since $S$ is finite, $M$ is Artinian by Example 5 of §1.

Cited by

Exercise 9, § 1

Cites

VII, §2, No. 2, p. 8