Exercise 4, § 1

Tag 0010, permanent. Cite it as [Bourbaki, Tag 0010].

Algebra, VIII, p. 14

Let $\rho : A\rightarrow B$ be a ring homomorphism.

a) If, as a right A-module, B is free and nonzero, and if the ring B is left Artinian (resp. left Noetherian), then the ring A is left Artinian (resp. left Noetherian).

b) Give an example where A is a commutative field, where B, viewed as a right vector space over A, is finite-dimensional, and where B is neither left Artinian nor left Noetherian (use Exercise 3).

Solution

This solution was written by gpt-5-6-mini and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

Part	Status	Reason
a	verified
b	verified

(a) Let $(e_i)_{i\in I}$ be a basis of the nonzero free right $A$ -module $B$ . Since $I$ is nonempty, the homomorphism $\rho$ is injective: indeed, if $\rho(a)=0$ , then $e_i\rho(a)=0$ for every $i$ , and the linear independence of the basis gives $a=0$ .

For every left ideal $J$ of $A$ , put

BJ=\sum_{j\in J}B\rho(j).

This is a left ideal of

B

. Moreover, since every element of

B

has a unique expression

\sum_{i\in I}e_i\rho(a_i)

with only finitely many

a_i

nonzero, we have

BJ=\left\{\sum_{i\in I}e_i\rho(a_i)\mathrel{}\middle|\mathrel{}a_i\in J,\ a_i=0\text{ for almost all }i\right\}.

Consequently,

BJ=BJ'

implies

J=J'

Suppose first that $B$ is left Artinian. By Definition 2 of § 1, every decreasing sequence of left ideals of $B$ is stationary. If

J_0\supset J_1\supset\cdots

is a decreasing sequence of left ideals of

A

, then

BJ_0\supset BJ_1\supset\cdots

is a decreasing sequence of left ideals of

B

, hence is stationary. The preceding injectivity then gives that

(J_n)

is stationary. Thus

A

is left Artinian.

If $B$ is left Noetherian, the same argument, with increasing sequences, shows that every increasing sequence of left ideals of $A$ is stationary. By Definition 2 of § 1, $A$ is therefore left Noetherian.

(b) By Exercise 3(a), choose a commutative field $K$ and a proper subfield $K'$ isomorphic to $K$ by an isomorphism $\varphi$ , with $K$ finite-dimensional over $K'$ . Let

C=K\times K

with multiplication

(x,y)(x',y')=(xx',xy'+y\varphi(x')).

By Exercise 3(a),

C

is left Artinian and left Noetherian but neither right Artinian nor right Noetherian.

Put

A=K,\qquad B=C^{o},

and embed

A

B

by the homomorphism corresponding in

C

a\longmapsto(a,0).

Since

C

is a two-dimensional left vector space over

K

for this scalar action,

B=C^{o}

is a two-dimensional right vector space over the commutative field

A=K

On the other hand, the left ideals of $B=C^{o}$ are precisely the right ideals of $C$ . Hence $B$ is neither left Artinian nor left Noetherian, by Exercise 3(a).

Cited by

Nothing.

Cites

Exercise 3