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The Krull dimension of a ring R is defined as the supremum of the lengths of chains of prime ideals contained in R. I heard that an integral extension over a ring R has the same Krull dimension as R, however, I don't really see why this is true.

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    $\begingroup$ You can find this in most commutative algebra books. Key words are going up and going down, if I recall correctly. $\endgroup$
    – Pedro Tamaroff
    Sep 8 '17 at 20:20
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The wiki article hints that an integral extension $R\subseteq S$ satisfies going-up, lying-over and incomparability, the Krull dimensions of $R$ and $S$ are the same.

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