Corollary 2
Tag 00R0, permanent. Cite it as [Bourbaki, Tag 00R0].
Algebra, VIII, p. 68, § 4. Semisimple Modules
Let M be an A-module. The following two conditions are equivalent:
(i) Every submodule of M admits a unique supplementary submodule.
(ii) The module M is the direct sum of a family of simple, pairwise nonisomorphic modules.
Suppose that M satisfies these conditions. Then, for every submodule N of M, there exists a unique subset J of I such that we have and every simple submodule of M is equal to one of the .
Conditions (i) and (ii) both imply that M is semisimple.
Suppose that condition (i) is satisfied. Let . By the equivalence of (i) and (iii) in Corollary 1, every submodule of is zero or equal to . Consequently, is zero or simple, and (ii) follows from the fact that M is the direct sum of the family .
Conversely, if condition (ii) is satisfied, then is zero or simple for every . If N is a submodule of M, then is equal to 0 or for every . Since we have (VIII, p. 66, Corollary), the submodule N has property (iii) of Corollary 1 and therefore admits a unique supplementary submodule in M. This proves that (ii) implies (i), as well as the last assertions of the corollary.
Let M be an A-module and S a simple A-module. Denote the opposite ring of the field End by D, and view S as an , D)-bimodule. Then Hom is a left vector space over D, and Hom is a right vector space over D. The dual of the left D-vector space Hom is a right vector space over D (II, §2, No. 3, p. 232, Definition 2). For every Hom, the mapping from Hom to Hom is a linear form on the left D-vector space Hom.
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
- II, §2, No. 3, p. 232, Definition 2 not in this corpus
- VIII, p. 66
- Corollary 1