Going down holds for integral extensions over integrally closed domains

Let $S:=B\setminus {\mathfrak{q}}_1$. By [L2], primes of $B_S=B_{{\mathfrak{q}}_1}$ correspond exactly to primes of $B$ contained in ${\mathfrak{q}}_1$. It is therefore enough to find a prime ideal of $B_{{\mathfrak{q}}_1}$ whose contraction to $A$ is ${\mathfrak{p}}_0$.

Let $b\in A\cap {\mathfrak{p}}_0{B}_{{\mathfrak{q}}_1}$. Write $b=y\mathrm{/}s$ with $y\in {\mathfrak{p}}_{0}B$ and $s\in B\setminus {\mathfrak{q}}_1$. Choose a representation $y={\sum }_{j=1}^m{a}_j{b}_j$ with $a_j\in {\mathfrak{p}}_0$ and $b_j\in B$. Because the extension is integral, [L1] says that each $b_j$ and $s$ is integral over $A$.

Let $M:=A[s,b_1,\dots ,b_m]$. By [L5], $M$ is module-finite over $A$. Since $y=\sum a_j{b}_j$ lies in ${\mathfrak{p}}_{0}B$, multiplication by $y$ sends $M$ into ${\mathfrak{p}}_{0}M$: for $m^{\mathrm{\prime }}\in M$, one has $y{m}^{\mathrm{\prime }}={\sum }_j{a}_j(b_j{m}^{\mathrm{\prime }})$ and each $b_j{m}^{\mathrm{\prime }}$ again lies in $M$. Choose $A$-generators $m_1,\dots ,m_r$ of $M$. Then there are coefficients $c_{ij}\in {\mathfrak{p}}_0$ with $y{m}_j={\sum }_i{c}_{ij}{m}_i$. Writing $C=(c_{ij})$, the adjugate identity [L6] applied to $y{I}_r-C$ yields a monic polynomial $P(T)=T^r+c_{r-1}{T}^{r-1}+\cdots +c_0$ with every $c_i\in {\mathfrak{p}}_0$ and $P(y)=0$.

Suppose $b\mathrm{\notin }{\mathfrak{p}}_0$. Let $K:=\mathrm{Frac}(A)$. Since $s$ is integral over $A$, [L4] gives a monic minimal polynomial $m_s(T)=T^d+d_{d-1}{T}^{d-1}+\cdots +d_0$ with coefficients in $A$. Because $y=bs$ and $P(y)=0$, dividing by $b^r$ in $K[T]$ gives a monic polynomial $Q(T)=T^r+\frac{c_{r-1}}{b}{T}^{r-1}+\cdots +\frac{c_0}{b^r}$ with coefficients in the localisation $A_{{\mathfrak{p}}_0}$ and satisfying $Q(s)=0$. Since $m_s$ is monic and divides every polynomial over $K$ that vanishes at $s$, it divides $Q$ in $K[T]$, hence also in $A_{{\mathfrak{p}}_0}[T]$.

Reduce that divisibility relation modulo ${\mathfrak{p}}_0{A}_{{\mathfrak{p}}_0}$. The image of $Q$ is $T^r$, so the image of $m_s$ is a monic divisor of $T^r$ in $\mathrm{Frac}(A\mathrm{/}{\mathfrak{p}}_0)[T]$. Therefore $\overline{m_s}(T)=T^d$, and every coefficient $d_i$ lies in ${\mathfrak{p}}_0$.

The relation $m_s(s)=0$ now shows that $s^d\in {\mathfrak{p}}_{0}B\subseteq {\mathfrak{q}}_1$. Because ${\mathfrak{q}}_1$ is prime, this forces $s\in {\mathfrak{q}}_1$, contradicting the choice of $s$. Hence $b\in {\mathfrak{p}}_0$. Therefore $A\cap {\mathfrak{p}}_0{B}_{{\mathfrak{q}}_1}={\mathfrak{p}}_0\mathrm{.}$

Let $C:=B_{{\mathfrak{q}}_1}\mathrm{/}{\mathfrak{p}}_0{B}_{{\mathfrak{q}}_1}$. Step 6.1 identifies $A\mathrm{/}{\mathfrak{p}}_0$ with a subring of $C$; because $A\mathrm{/}{\mathfrak{p}}_0$ is a domain, its nonzero elements form a multiplicative subset $T\subseteq C$. The zero ideal of $C$ is disjoint from $T$, so [L3] gives a prime ideal $\mathfrak{n}$ of $C$ disjoint from $T$. Its inverse image ${\mathfrak{q}}_0^{\mathrm{\prime }}$ in $B_{{\mathfrak{q}}_1}$ contains ${\mathfrak{p}}_0{B}_{{\mathfrak{q}}_1}$, and the disjointness from $T$ says exactly that ${\mathfrak{q}}_0^{\mathrm{\prime }}\cap A={\mathfrak{p}}_0$.

By [L2], the prime ${\mathfrak{q}}_0^{\mathrm{\prime }}$ of $B_{{\mathfrak{q}}_1}$ is the localisation of a unique prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ of $B$. Its contraction to $A$ is ${\mathfrak{p}}_0$ by step 7.1. Therefore ${\mathfrak{q}}_0$ is the required prime below ${\mathfrak{q}}_1$.