Bourbaki

Exercise 15, § 8

Algebra, X, p. 203

Let σ\sigma be an automorphism of the ring A. a) We denote by Aσ[X]A_\sigma[X] the ring defined in VIII, § 1, No. 3. Now prove that

dh(Aσ[X])=dh(A)+1. \mathrm{dh}(A_\sigma[X]) = \mathrm{dh}(A) + 1 .
(The inequality dh(Aσ[X])≥dh(A)+1\mathrm{dh}(A_\sigma[X]) \geq \mathrm{dh}(A) + 1 follows from Exercise 14, b); for the opposite inequality, take inspiration from the proof of X, p. 142, Lemma 3, a).) b) We denote by Aσ[[X]]A_\sigma[[X]] the ring defined in IV, § 4, Exercise 8. Now prove that if A is noetherian, one has dh(Aσ[[X]])=dh(A)+1\mathrm{dh}(A_\sigma[[X]]) = \mathrm{dh}(A) + 1 (use Exercise 14).

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