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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?

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    Please 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

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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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    Thank 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