Bourbaki

Exercise 5, Appendix 1

Tag 00PQ, permanent. Cite it as [Bourbaki, Tag 00PQ].

Algebra, VIII, p. 443

Bourbaki marks this one as harder than the rest.

Suppose that every pseudomodule M over A is the direct sum of AM and the sub-pseudomodule of M consisting of the elements annihilated by A. Prove that A is unital. (Taking M=A~sM =\widetilde{A}_s, prove that A admits a right unit element uu. Taking M=AsM = A_s, prove that we have Ax≠0Ax\not= 0 for every nonzero element xx of A. Conclude that uu is also a left unit element.)

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Transcribed from the printed volume. Not translated.