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.

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.
