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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 $k$. –  user18119 Oct 29 '12 at 10:21

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up vote 2 down vote accepted

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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In Introduction to commutative algebra There is used a lot of properties of dimensions of local rings. Where is there simple proof, using only Noether normalization? –  user46336 Oct 29 '12 at 18:11
    
Which new reference? Can you advise me a some book or link? –  user46336 Oct 29 '12 at 18:29
    
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 minimal prime ideal, then the transcendence degree decreases exactly by 1. –  user46336 Oct 29 '12 at 20:11

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