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
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abstract-algebra
algebraic-geometry
commutative-algebra
ring-theory
krull-dimension
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2Please define the trancendence degree of an algebra (I guess it is that of its field of fractions). It is not true in general: take for $R$ a non-algebraic field extension of $k$. – 2012-10-29
1 Answers
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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.)
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0Thank you, yes! It's obvious, that the thanscendence degree is not lower, than Krull. The thing to be proved is that if we factorize by the $m$inimal prime ideal, then the transcendence degree decreases exactly by 1. – 2012-10-29