Bourbaki

Exercise 5, § 2

Tag 002P, permanent. Cite it as [Bourbaki, Tag 002P].

Algebra, VIII, p. 39

Let A be a ring.

a) An element ee of A is idempotent if and only if the mapping x↦xex\mapsto xe is a projector of the left A-module AsA_s (resp. the mapping x↦exx\mapsto ex is a projector of the right A-module AdA_d).

b) Let e1,e2e_1, e_2 be idempotents in the ring A. Prove that the group HomA(Ae1,Ae2)_A(Ae_1,Ae_2) is isomorphic to e1A∩Ae2=e1Ae2e_1A\cap Ae_2=e_1Ae_2 and that the ring EndA(Ae1)_A(Ae_1) is isomorphic to the ring e1Ae1e_1Ae_1.

c) Let e1,e2e_1, e_2 be idempotents in A. Prove that the following properties are equivalent:

(i) Ae1=Ae2Ae_1= Ae_2.

(ii) e1e2=e1e_1e_2=e_1 and e2e1=e2e_2e_1=e_2.

(iii) (1−e1)A=(1−e2)A(1-e_1)A = (1-e_2)A.

When they are satisfied, there exists an invertible element a∈Aa\in A such that e2=e_2= ae1a−1ae_1a^{-1}. d) Let e1,e2e_1, e_2 be two idempotents in A. Prove that the following properties are equivalent:

(i) The left A-modules Ae1Ae_1 and Ae2Ae_2 are isomorphic.

(ii) The right A-modules e1Ae_1A and e2Ae_2A are isomorphic.

(iii) There exist elements x,yx, y of A such that xy=e1xy=e_1 and yx=e2yx=e_2.

When they are satisfied, we say that e1e_1 and e2e_2 are equivalent idempotents.

e) If idempotents in A are equivalent and belong to the center of A, then they are equal.

f ) Let (ei)i∈I(e_i)_{i\in I} and (fi)i∈I(f_i)_{i\in I} be finite families of idempotents in A such that eiej=0e_ie_j= 0 and fifj=0f_if_j= 0 for i∈I,j∈I,i≠ji\in I,j\in I,i\not=j and that eie_i is equivalent to fif_i for every i∈Ii\in I. Then e=∑i∈Ieie=\sum_{i\in I}e_i and f=∑i∈Ifif=\sum_{i\in I}f_i are equivalent idempotents in A.

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Transcribed from the printed volume. Not translated.