Integral closure commutes with étale base change

Because $E$ is flat over $R$, [F1] makes $E{\otimes }_{R}A\to E{\otimes }_{R}B$ injective, and we identify its source with its image. Every element of $A$ is integral over the image of $R$, so every element of the algebra $E{\otimes }_{R}A$ is integral over $E$: each is a finite sum of products of elements $1\otimes a$ satisfying monic equations over $E$, and integral elements form a subring by [F1]. This proves one inclusion.

For the converse first take a monic polynomial $P\in R[T]$ of degree $d\ge 1$, put $C=R[T]\mathrm{/}(P)$ and $D=B[T]\mathrm{/}(P)$, and let $c\in D$ be integral over $C$. As $C$ is finite free over $R$ with basis $1,T,\dots ,T^{d-1}$, it is integral over $R$; transitivity of integrality makes $c$ integral over $R$. Write uniquely $c={\sum }_{j<d}{b}_j{T}^j$ with $b_j\in B$. The claim needed for étale charts is that every coefficient of $P^{\mathrm{\prime }}(T)c$ reduced modulo $P$ belongs to $A$.

Construct the splitting algebra $B^{\mathrm{\prime }}$ of $P$ by adjoining one formal root at a time: quotient successively by the current monic polynomial, divide by $(T-{\alpha }_i)$, and continue. Each quotient is free over its predecessor with a basis containing $1$, so the composite $B\to B^{\mathrm{\prime }}$ is injective. In $B^{\mathrm{\prime }}[T]$ one has $P(T)={\prod }_{i=1}^d(T-{\alpha }_i)$, with each ${\alpha }_i$ integral over $R$. For every polynomial $h(T)$ of degree less than $d$ the identity $P^{\mathrm{\prime }}(T)h(T)\equiv {\sum }_{i=1}^{d}h({\alpha }_i){\prod }_{j\mathrm{\ne }i}(T-{\alpha }_j)(\mathrm{m}\mathrm{o}\mathrm{d}P(T))$ holds without distinct-root or denominator assumptions: it is an identity over the polynomial ring in independent formal roots, where it follows by checking the degree-less-than-$d$ interpolation basis, and polynomial specialization preserves it. Apply it to the representative of $c$. Since $c$ is integral over $R$, every evaluation $c({\alpha }_i)$ is integral over $R$; each coefficient on the right is a sum of products of these evaluations and the integral roots ${\alpha }_i$, hence integral over $R$ by [F1]. The left side has its reduced coefficients in the injected subring $B\subseteq B^{\mathrm{\prime }}$, so those coefficients lie in $A$ by the definition of $A$. Thus $P^{\mathrm{\prime }}c\in A[T]\mathrm{/}(P)$ inside $D$.

Now let $C_g=(R[T]\mathrm{/}(P){)}_g$ be a standard étale algebra, so $P^{\mathrm{\prime }}$ is a unit in $C_g$ by [F2], and let $b\in D_g$ be integral over $C_g$. Clearing the finitely many denominators in a monic equation for $b$ gives an integer $N$ and an element $c=g^{N}b\in D$ integral over $C$; this is the localization property [F3] applied to the relative integral closure. Step 2.1 gives $P^{\mathrm{\prime }}c\in A[T]\mathrm{/}(P)$. Since $g$ and $P^{\mathrm{\prime }}$ are units in $C_g$, both $g^{-N}$ and $(P^{\mathrm{\prime }}{)}^{-1}$ belong to $C_g$, so $b=g^{-N}(P^{\mathrm{\prime }}{)}^{-1}(P^{\mathrm{\prime }}c)$ lies in the image of $C_g{\otimes }_{R}A\to C_g{\otimes }_{R}B$. Thus the converse inclusion holds for every standard étale chart.

For general $E$, use the standard-étale affine neighbourhoods of [F2] and the base localization compatibility [F3]. On each such neighbourhood, step 3.1 proves that any element integral over $E$ lies in the localized image of $E{\otimes }_{R}A$. For each integral element $b$, its class in the quotient module $(E{\otimes }_{R}B)\mathrm{/}(E{\otimes }_{R}A)$ vanishes after localization at every prime on that cover, so that class is zero globally. Together with step 1.1 this proves equality. If $E=0$ or $B=0$, both sides are zero and the same argument is vacuous. AC is inherited only through [F2] and [F3]; the finite splitting construction uses no further choice. The cited local standard-étale theorem supplies [F2], completing the general case.