Bourbaki

Proposition 5

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

Commutative Algebra, chapter VI, § 8. Extensions of a valuation to an algebraic extension

Let KK be a field, v a valuation on KK, AA its ring, m its ideal, L a finite extension of KK of degree nn, BB the integral closure of AA in LL and (vi′)1≤i≤s(v'_i)_{1 \leq i \leq s} a complete system of extensions of vv to LL. Then

[B/mB:A/m]=∑i=1sε(vi′/v)f(vi′/v). [B/mB : A/m] = \sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v).

Let AiA_i be the ring of vi′v'_i; then Ai=BmiA_i = B_{m_i} where mim_i runs through the family of maximal ideals of BB (no. 3, Remark). Let qiq_i be the saturation of mBmB with respect to mim_i (Chapter II, § 2, no. 4). By Chapter V, Corollary 3 to Proposition 1, no. 1, § 2, the canonical homomorphism B/mB→∏i=1sB/qiB/mB \to \prod_{i=1}^s B/q_i is an isomorphism and mim_i is the only maximal ideal of B containing qiq_i. Hence B/qiB/q_i is canonically isomorphic to (B/qi)mi(B/q_i)_{m_i} (Chapter II, § 3, no. 3, Proposition 8), that is to

Bmi/mBmi=Ai/mAi. B_{m_i}/mB_{m_i} = A_i/mA_i.
Therefore there is a canonical isomorphism B/mB→∏i=1sAi/mAiB/mB \to \prod_{i=1}^s A_i/mA_i, whence the result by virtue of Proposition 4 of no. 4.

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.