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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.

McNuggets666
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    You can find this in most commutative algebra books. Key words are going up and going down, if I recall correctly. – Pedro Sep 08 '17 at 20:20
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    Was looking for it in Atiyah-Macdonald and couldn't find it, not even as an exercise... I guess they find too obvious :) For future reference, we can combine Corollary 4.9 (incomparability of distinct primes lying above the same prime) and Theorem 5.11 (going-up) to get the result. – Anakhand Jan 28 '24 at 12:54
  • Here is a proof. – Bowei Tang Jun 08 '25 at 13:12

1 Answers1

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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.

rschwieb
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