Corollary 1

Tag 003H, permanent. Cite it as [Bourbaki, Tag 003H].

Algebra, VIII, p. 48, § 3. Simple Modules

Keep the assumptions of Theorem 1, and suppose, moreover, that the algebra A is commutative. Then M has dimension 1 over 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.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

Cited by

§ 21. Linear Representations of Finite Groups

Cites

Theorem 1