# Why does an integral extension over a ring has the same Krull dimension as the ring?

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.

• You can find this in most commutative algebra books. Key words are going up and going down, if I recall correctly. Sep 8 '17 at 20:20

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.