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?
2026-03-29 15:14:51.1774797291
Krull dimension and transcendence degree
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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.)