Bourbaki

Example

Tag 05HC, permanent. Cite it as [Bourbaki, Tag 05HC].

Algebra, chapter I, § 8. Rings

(1) Let AA be a ring. The set AA is a two-sided ideal of AA; so is the set consisting of 0, which is called the zero ideal and sometimes denoted by 0 or (0)(0) instead of {0}\{0\}.

(2) For every element aa of AA, the set A.aA.a of left multiples of aa is a left ideal; similarly the set a.Aa.A is a right ideal. When aa is in the centre of AA, A.a=a.AA.a = a.A; this ideal is called the principal ideal generated by aa and is denoted by (a)(a). (a)=A(a) = A if and only if aa is invertible.

(3) Let MM be a subset of AA. The set of elements x∈Ax \in A such that xy=0xy = 0 for all y∈My \in M is a left ideal of AA called the left annihilator of MM. The right annihilator of MM is defined similarly.

(4) Every intersection of left (resp. right, two-sided) of AA is a left (resp. right, two-sided) ideal. Given a subset XX of AA, there thus exists a smallest left (resp. right, two-sided) ideal containing XX; it is called the left (resp. right, two-sided) ideal generated by XX.

Let aa be a left ideal of AA. The conditions 1∉a,a≠A1 \notin a, a \neq A are obviously equivalent.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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Cites

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Transcribed from the printed edition, 1998, Springer. Not translated.