Bourbaki

Exercise 5, § 8

Commutative Algebra, VI, 462

Let K be a field, A a valuation ring of K and L a finite algebraic extension of K. Let A ⊃ A be another valuation ring of K; let A'_i (1 ≤ i ≤ m) be the valuation rings of L such that A'_i ∩ K = A' and let k'_i be their respective residue fields. Let k be the residue field of A' and A'' the valuation ring of K the canonical image of A; finally let A''_{ij} (1 ≤ j ≤ n_i) be the valuation rings of k'_i such that A''_{ij} ∩ k = A'' and A_{ij} the inverse images in A'_i of the A''_{ij} (1 ≤ i ≤ m, 1 ≤ j ≤ n_i).

(a) Show that the A_{ij} are valuation rings of L which are distinct and such that A_{ij} ∩ K = A and every valuation ring B of L such that B ∩ K = A is equal to one of the A_{ij}.

(b) Show that

e(Aij/A)=e(Ai′/A′)e(Aij′′/A′′) e(A_{ij}/A) = e(A'_i/A')e(A''_{ij}/A'')
and
f(Aij/A)=f(Aij′′/A′′). f(A_{ij}/A) = f(A''_{ij}/A'').

Solution

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Transcribed from the printed volume. Not translated.