An integral domain extension can fail going down when the base is not normal

The ring $S$ is a domain, and $R$ is a subring of it, so $R$ is also a domain. Moreover $x$ satisfies the monic equation $T^2-T+x(1-x)=0$ with coefficient $x(1-x)\in R$, and $S=R[x]$ because $y\in R$ already. Hence $S$ is integral over $R$.

The ideals $P_0$ and $P$ are prime because they are contractions of the prime ideals $(x)$ and $(x,y)$ of $S$, and they are distinct because $y\in P\setminus P_0$. The ideal $Q=(1-x,y)$ is prime in $S$, and its contraction to $R$ is $P$, since mod $Q$ one has $x=1$ and $y=0$, so $x(1-x)$, $y$, and $xy$ all vanish.

The base ring is not integrally closed: the element $x\in S$ is integral over $R$ by step 1.1, but $x\mathrm{\notin }R$. Indeed, if $x\in R$, then setting $y=0$ would express $x$ as a polynomial in $x(1-x)$; evaluating at $x=0$ and $x=1$ would then give the same value on both inputs, impossible because $x$ takes the values $0$ and $1$.

Suppose $Q_0\subseteq Q$ were a prime ideal of $S$ with $Q_0\cap R=P_0$. Because $Q=(1-x,y)$ does not contain $x$, neither does $Q_0$. But $x(1-x)\in P_0\subseteq Q_0$ and $xy\in P_0\subseteq Q_0$, so primality of $Q_0$ and $x\mathrm{\notin }{Q}_0$ force $1-x\in Q_0$ and $y\in Q_0$. Hence $Q=(1-x,y)\subseteq Q_0$, and therefore $Q_0=Q$. This contradicts $Q_0\cap R=P_0$ because step 2.1 showed $Q\cap R=P\mathrm{\ne }{P}_0$.

Thus the integral extension of domains $R\subseteq S$ has a prime chain $P_0⊊P$ and a prime $Q$ over $P$ with no prime below $Q$ lying over $P_0$. So the normality hypothesis in [L1] is essential.