Bourbaki

Exercise 15, § 5

Commutative Algebra, II

Bourbaki marks this one as harder than the rest.

Let BB be an Azumaya A-algebra (exer. 14). a) Now prove that the center of BB is identical with the subring AA of BB and is a direct factor of the a-module BB (cf. Chap. I, § 2, exer. 21, and Chap. II, § 5, exer. 4). b) Now prove that, for every two-sided ideal bb of B, one has b=(b∩A)Bb = (b \cap A)B. (First note that bb is stable under every endomorphism of the a-module B, and use the existence of a projector of B onto A.)

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.