Bourbaki

Example

Tag 05H8, permanent. Cite it as [Bourbaki, Tag 05H8].

Algebra, chapter I, § 8. Rings

(1) Let A be a ring. It is immediately seen that the mapping n↦n⋅1n \mapsto n \cdot 1 of Z\mathbf{Z} into A is the unique homomorphism of Z\mathbf{Z} into A. In particular, the identity mapping of Z\mathbf{Z} is the unique endomorphism of the ring Z\mathbf{Z}.

In particular, take A to be the endomorphism ring of the additive group Z\mathbf{Z} (no. 3, Example VI). The mapping n↦n⋅1n \mapsto n \cdot 1 of Z\mathbf{Z} into A is an isomorphism of Z\mathbf{Z} onto A by the very construction of multiplication in Z\mathbf{Z} (§ 2, no. 6).

(2) Let aa be an invertible element of a ring A. The mapping x↦axa−1x \mapsto axa^{-1} is an endomorphism of A for

a(x+y)a−1=axa−1+aya−1, a(x + y)a^{-1} = axa^{-1} + aya^{-1},
a(xy)a−1=(axa−1)(aya−1). a(xy)a^{-1} = (axa^{-1})(aya^{-1}).

It is bijective, for the relation x′=axa−1x' = axa^{-1} is equivalent to x=a−1x′ax = a^{-1}x'a. It is therefore an automorphism of the ring A, called the inner automorphism associated with aa.

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

Cited by

Nothing.

Cites

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