Integral closure is unchanged across an integral intermediate domain

Suppose first that $z$ is integral over $A$. By [L1] there is a monic polynomial $f\in A[X]$ with $f(z)=0$. Since $A\subseteq B$, the same polynomial, viewed in $B[X]$, is monic and has the same root $z$; by [L1] again, $z$ is integral over $B$.

Conversely assume that $z$ is integral over $B$. By [L1] there are an integer $n\ge 1$ and coefficients $b_0,\dots ,b_{n-1}\in B$ with $z^n+b_{n-1}{z}^{n-1}+\cdots +b_{1}z+b_0=0.$

Each $b_i$ lies in $B$, and $B$ is integral over $A$, so every $b_i$ is integral over $A$ by [L2]. Let $C:=A[b_0,\dots ,b_{n-1}]$ be the subring of $L$ generated over $A$ by these coefficients. By [L3], applied to the ring extension $A\subseteq L$ (legitimate by [L5]), the elements of $L$ integral over $A$ form a subring of $L$; it contains $A$ and every $b_i$, hence contains the subring $C$ these elements generate. Therefore the inclusion $A\to C$ is an integral ring map.

The equation of step 1.2 has all its coefficients in $C$, so by [L1] the element $z$ is integral over $C$. Applying [L3] to the ring extension $C\subseteq C[z]$ (again $C\mathrm{\ne }0$ by [L5]) shows that the elements of $C[z]$ integral over $C$ form a subring containing $C$ and $z$, hence containing the subring $C[z]$ that they generate; therefore the inclusion $C\to C[z]$ is an integral ring map.

By steps 2.1 and 3.1 the maps $A\to C$ and $C\to C[z]$ are both integral, so the composite inclusion $A\to C[z]$ is integral by [L4]; every element of $C[z]$ is thus integral over $A$ by [L2], and in particular $z$ is integral over $A$. Together with step 1.1 this proves both directions of the equivalence.