Relative integral closure under elementary etale change

Flatness of $E$ preserves the injection $C\hookrightarrow B$. The image of $E{\otimes }_{A}C$ is integral over $E$, since each of its elements uses finitely many integral elements of $C$ and sums and products of integral elements remain integral by [F2]. For the opposite inclusion it suffices to work on the monic principal-open charts of [F1]; an element belongs to a submodule if its class in the quotient vanishes locally on a covering family.

Let $P\in A[T]$ be monic of degree $d>0$, put $D=A[T]\mathrm{/}(P)$, and take $c\in B[T]\mathrm{/}(P)$ integral over $D$. Since $D$ is finite free over $A$, transitivity makes $c$ integral over $A$. Write $c={\sum }_{i<d}{b}_i{T}^i$. Form a splitting algebra over $B$ by adjoining a root of $P$, dividing by its monic linear factor, then adjoining a root of the quotient and continuing. Every extension is finite free with a basis containing $1$, so $B$ injects into the splitting algebra, and its roots ${\alpha }_1,\dots ,{\alpha }_d$ are integral over $A$. The polynomial identity $P^{\mathrm{\prime }}(T)c(T)\equiv {\sum }_{i=1}^{d}c({\alpha }_i){\prod }_{j\mathrm{\ne }i}(T-{\alpha }_j)(\mathrm{m}\mathrm{o}\mathrm{d}P(T))$ holds universally: over independent formal roots, interpolation proves it after inverting the Vandermonde determinant, and injectivity of that localization in the universal polynomial domain proves the original polynomial identity. Specialization therefore permits repeated roots. Each $c({\alpha }_i)$ is integral over $A$, because evaluation carries its monic equation to a monic equation. Every coefficient on the right is integral over $A$. The degree-less-than-$d$ remainder on the left has coefficients in the injected ring $B$, so each of those coefficients belongs to $C$. Thus $P^{\mathrm{\prime }}c\in C[T]\mathrm{/}(P)$.

On the chart $E=D_h$ of [F1], let $b\in (B[T]\mathrm{/}(P){)}_h$ be integral over $E$. Localization compatibility in [F2] gives some $N$ such that $c=h^{N}b$ lies in $B[T]\mathrm{/}(P)$ and is integral over $D$. By step 1.2, $P^{\mathrm{\prime }}c$ lies in $C[T]\mathrm{/}(P)$. Both $h$ and $P^{\mathrm{\prime }}$ are invertible in $E$, whence $b=h^{-N}(P^{\mathrm{\prime }}{)}^{-1}(P^{\mathrm{\prime }}c)$ belongs to $E{\otimes }_{A}C$. Covering by these charts proves the opposite inclusion. Degree-zero monic charts are empty and cause no exception; the zero algebras are likewise harmless. Together with step 1.1 this proves the equality. AC is inherited through [F1]; the finite splitting-algebra construction adds no further choice.