# Krull dimension and transcendence degree

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