Bourbaki

Theorem 1

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

Algebra, VIII, p. 47, § 3. Simple Modules

Let K be an algebraically closed commutative field, A a K-algebra, and M a simple A-module. Suppose that the dimension of M as a vector space over K is finite or, more generally, strictly less than the cardinal of K. Then the endomorphism ring of the A-module M consists of the homotheties αM\alpha_M with α∈K\alpha \in K.

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 dimKM=_KM = (dimEM_EM)[E:K][E : K] by Proposition 25 of II, §1, No. 13, p. 222, hence dimKM⩾[E:K]_KM\geqslant [E : K]. Since dimKM<_KM< Card(K) by assumption, the equality E=K⋅1ME = K\cdot 1_M is then a consequence of the following lemma.

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

Cites

Transcribed from the printed edition, 2023, Springer Nature. Not translated.