Going down theorem fails
HINT $\ $ Put $T = k[x],\ x_1 = x(x-1),\ R = k[x_1,xx_1].$ Note that $T[z]$ is the integral closure of $R[z]$ since $T$ is the integral closure of $R.$ To show that GD fails for $R[z]\subset T[z]$ consider the prime ideals $P = x_1\:(z-x)\ \subset R[z]$ and $Q = (x-1,z)\subset T[z].$ Note that $Q$ lies over the prime $Q\cap R[z] = (x_1,xx_1,z)\supseteq P$ but $T[z]$ has no prime in $Q$ lying over $P.$
Note that the GD theorem doesn't apply since $R$ is not integrally closed. Below is further discussion from Matsumura's Commutative Algebra, p. 32.
