Bourbaki

Definition 1

Tag 04T2, permanent. Cite it as [Bourbaki, Tag 04T2].

Commutative Algebra, chapter VI, § 3. Valuations

Let CC be a (not necessarily commutative) ring and Γ\Gamma a totally ordered commutative group written additively. A valuation on CC with values in Γ\Gamma is any mapping v:C→Γ∞v : C \to \Gamma_\infty which satisfies the following conditions:

(VL_I) v(xy)=v(x)+v(y)v(xy) = v(x) + v(y) for x∈C,y∈Cx \in C, y \in C. (VL_{II}) v(x+y)≥inf⁡(v(x),v(y))v(x + y) \geq \inf(v(x), v(y)) for x∈C,y∈Cx \in C, y \in C. (VL_{III}) v(1)=0v(1) = 0 and v(0)=+∞v(0) = +\infty.

If CC has no divisor of zero other than 0, the unique mapping v0v_0 of CC to Γ∞\Gamma_\infty such that v0(x)=0v_0(x) = 0 for x≠0x \neq 0 and v0(0)=+∞v_0(0) = +\infty is a valuation, called the improper valuation on CC. If z∈Cz \in C is such that zn=1z^n = 1 for some integer n≥1n \geq 1, then, by (VL_I), nv(z)=v(zn)=0nv(z) = v(z^n) = 0 and hence v(z)=0v(z) = 0 for every valuation vv on CC, since Γ\Gamma is a totally ordered group. In particular v(−1)=0v(-1) = 0, whence v(−x)=v(x)v(-x) = v(x) for all x∈Cx \in C. Moreover, it follows from (VL_I) that v(xy)=v(yx)v(xy) = v(yx) for all x,yx, y in CC. If xx is invertible in CC, then v(x−1)=−v(x)v(x^{-1}) = -v(x).

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

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Transcribed from the printed volume. Not translated.