What is the simplest proof of the fact that an integral algebra $R$ over a field $k$
has the same Krull dimension as transcendence degree $\operatorname{trdeg}_k R$?
Is it possible to use only Noether normalization theorem?
Krull Dimension and Transcendence Degree in Algebraic Geometry
abstract-algebraalgebraic-geometrycommutative-algebrakrull-dimensionring-theory
Best Answer
R. Ash, A Course in Commutative Algebra, proof of Theorem 5.6.7 uses Noether normalization and few obvious remarks on integral extensions. (However, see QiL's comment.)