Bourbaki

Lemma 2

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

Commutative Algebra, chapter VI, § 10. Extensions of a valuation to a transcendental extension

Let H be an isolated subgroup of a totally ordered commutative group G. If H is a direct factor of G, the ordered group G is isomorphic to the group (G/H)×H(G/H) \times H ordered lexicographically.

Let jj be an isomorphism of (G/H)×H(G/H) \times H onto G such that j(0,x)=xj(0, x) = x for all x∈Hx \in H and j(y,x)j(y, x) belongs to the coset of H. As (G/H)×H(G/H) \times H is totally ordered, it amounts to showing that jj is increasing (Set Theory, Chapter III, § 1, no. 12, Proposition 11). Let (y,x)(y, x) be an element ≥0\geq 0 of (G/H)×H(G/H) \times H ordered lexicographically. If y>0y > 0, the coset of H containing j(y,x)j(y, x) is an element >0> 0, whence j(y,x)>0j(y, x) > 0, for, otherwise, y≤0y \leq 0 (§ 4, no. 2, Proposition 3). If y=0y = 0 and x≥0x \geq 0, then j(y,x)=x≥0j(y, x) = x \geq 0. Hence jj is certainly increasing.

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

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