Bourbaki

Example (6)

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

Commutative Algebra, chapter VI, § 3. Valuations

Let Γ be a totally ordered group and k a field. Let Γ_+ be the monoid of positive elements of Γ and C the algebra of Γ_+ over k. By definition, C has a basis (x_\alpha)_{\alpha \in \Gamma_-} over k whose multiplication table is x_\alpha x_\beta = x_{\alpha + \beta}. If x=∑aαxαx = \sum a_\alpha x_\alpha is a non-zero element of C, we write v(x)=inf⁡aα≠0(a)v(x) = \inf_{a_\alpha \neq 0} (a) and v(0)=+∞v(0) = +\infty; it is immediately verified that the mapping vv of CC to Γ∞\Gamma_\infty satisfies conditions (VL_I) and (VL_{II}) of no. 1 and that CC is an integral domain. Let KK be the field of fractions of CC and ww the valuation on KK which extends vv (Proposition 4, no. 2). As every element of Γ\Gamma is the difference of two positive elements, ww admits Γ\Gamma as order group. Let AA be the ring of ww and mm its maximal ideal; we shall show that AA is the direct sum of mm and kk (identified with k⋅1k \cdot 1), which will prove that the residue field of ww is isomorphic to kk. Clearly m∩k=(0)m \cap k = (0). On the other hand, denoting by pp the ideal of CC generated by the xax_a where a>0a > 0, every element xx of valuation 0 in KK can be written in the form (a+y)/(b+z)(a + y)/(b + z) where a∈k∗,b∈k∗,y∈pa \in k^*, b \in k^*, y \in p and z∈pz \in p; then

x=ab−1+(by−az)b−1(b+z)−1 x = ab^{-1} + (by - az)b^{-1}(b + z)^{-1}

whence w(x−ab−1)>0w(x - ab^{-1}) > 0 and x≡ab−1(modm)x \equiv ab^{-1} \pmod{m}; this shows our assertion.

If Γ=Z×Z\Gamma = \mathbf{Z} \times \mathbf{Z}, then K=k(X,Y)K = k(X, Y) and the above construction then provides valuations on k(X,Y)k(X, Y) which are improper on kk, whose order group is Z×Z\mathbf{Z} \times \mathbf{Z} and whose residue field is kk. These valuations depend on the order structure chosen on Z×Z\mathbf{Z} \times \mathbf{Z}. For example, Z×Z\mathbf{Z} \times \mathbf{Z} can be given the lexicographic ordering. Or indeed, for an irrational number aa, Z×Z\mathbf{Z} \times \mathbf{Z} may be identified with a subgroup of R\mathbf{R} under the homomorphism (m,n)↦m+nα(m, n) \mapsto m + n\alpha (a homomorphism which is injective since aa is irrational) and given the ordering induced by that on R\mathbf{R}.

Other constructions of valuations using Proposition 4 of no. 2 will be described in § 10.

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

Cited by

Cites

Transcribed from the printed volume. Not translated.