Bourbaki

Example

Tag 04U6, permanent. Cite it as [Bourbaki, Tag 04U6].

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

Let kk be a field,

E=k(X)andK=k(X,Y)=E(Y) E = k(X) \quad \text{and} \quad K = k(X, Y) = E(Y)

(X,Y(X, Y indeterminates). Let B=E[Y](Y)B = E[Y]_{(Y)} be the valuation ring of KK defined by the extremal element YY of the principal ideal domain E[Y]E[Y] (§ 1, no. 4, Proposition 3). The residue field κ(B)\kappa(B) is canonically identified with E[Y]/(Y)=EE[Y]/(Y) = E. Similarly, let A′=k[X](X)A' = k[X]_{(X)} be the valuation ring of E=k(X)E = k(X) defined by the extremal element XX of k[X]k[X]. Denoting by hh, the place of EE associated with BB and writing A=hB−1(A′)A = h_B^{-1}(A'), a valuation ring AA of KK is defined which is contained in BB and κ(A)=κ(A′)=k\kappa(A) = \kappa(A') = k. The canonical place hh: K→kK \to k can be described as follows: iff (X,Y)(X, Y) is an element of KK, then we first put Y=0Y = 0 inf (which gives an element of E~=k(X)∗\tilde{E} = k(X)^*), then X=0X = 0 in the result obtained. The groups ΓA′\Gamma_{A'} and ΓB\Gamma_B are canonically isomorphic to Z\mathbf{Z} (§ 3, no. 4, Example 4). It is not difficult to show (cf. § 10, no. 2, Lemma 2) that the group ΓA\Gamma_A is isomorphic to the lexicographic product Z×Z\mathbf{Z} \times \mathbf{Z} and that the valuation vAv_A is equivalent to the valuation defined in § 3, no. 4, end of Example 6.

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 volume. Not translated.