Remark

Tag 04U5, permanent. Cite it as [Bourbaki, Tag 04U5].

Commutative Algebra, chapter VI, § 4. The height of a valuation

Under the above hypotheses, let $f$ denote the canonical homomorphism of $B$ onto $\kappa(B)$ and $A' = f(A)$ ; it is a valuation ring of $\kappa(B)$ (Proposition 2, no. 1). Then $f^{-1}(\kappa(B)^*) = U(B), f^{-1}(A') = A, f^{-1}(m(A')) = m(A)$ , hence

f^{-1}(U(A')) = U(A).

Then there is a canonical isomorphism of $U(B)/U(A)$ onto $\kappa(B)^*/U(A') = \Gamma_{A'}$ . The exact sequence

0 \to U(B)/U(A) \to \Gamma_A \to \Gamma_B \to 0

then gives an exact sequence

0 \to \Gamma_{A'} \to \Gamma_A \to \Gamma_B \to 0.

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

Cited by

Nothing.

Cites

Proposition 2