Bourbaki

Proposition 3

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

Algebra, chapter I, § 9. Fields

*Let B\mathbf{B} be a non-zero ring and A\mathbf{A} a commutative subring of B\mathbf{B} such that every non-zero element of A\mathbf{A} is invertible in B\mathbf{B}. (a) A\mathbf{A} is an integral domain. (b) *Let A′\mathbf{A}' be the field of fractions A\mathbf{A}. The canonical injection of A\mathbf{A} into B\mathbf{B} can be extended uniquely to an isomorphism ff of A′\mathbf{A}' onto a subfield of B\mathbf{B}. (c) The elements of f(A′)f(\mathbf{A}') are the xy−1xy^{-1} where x∈A, y∈A, y≠0x \in \mathbf{A},\ y \in \mathbf{A},\ y \neq 0.

Assertion (a) is obvious. The canonical injection of A\mathbf{A} into B\mathbf{B} extends uniquely to a homomorphism ff of A′\mathbf{A}' into B\mathbf{B} (§ 8, no. 12, Theorem 5). Assertion (b) then follows from no. 1, Theorem 2. The elements of A′\mathbf{A}' are the fractions x/yx/y with x∈A, y∈A, y≠0x \in \mathbf{A},\ y \in \mathbf{A},\ y \neq 0 and f(x/y)=xy−1f(x/y) = xy^{-1}, whence (c).

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.