Finite separable integral closures over normal Noetherian domains are module-finite

Choose an $F$-basis $e_1,\dots ,e_n$ of $L$. For each $i$, choose a monic polynomial for $e_i$ over $F$ and one nonzero denominator $d_i\in R$ clearing all of its coefficients. Then $(d_i{e}_i{)}^m+d_i{c}_{m-1}(d_i{e}_i{)}^{m-1}+\cdots +d_i^m{c}_0=0$ for suitable $c_j\in F$, so $d_i{e}_i$ is integral over $R$ and therefore lies in $A$. Replacing each $e_i$ by $d_i{e}_i$ preserves the $F$-basis, so we may assume from the start that $e_1,\dots ,e_n\in A$. By [L1], there is a trace-dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ with ${\mathrm{Tr}}_{L\mathrm{/}F}(e_i{e}_j^{\ast })={\delta }_{ij}$. Clearing denominators in the coordinates of the $e_j^{\ast }$ relative to the basis $e_i$, choose $0\mathrm{\ne }c\in R$ such that $c{e}_j^{\ast }\in {\sum }_{i}R{e}_i\subseteq A$ for every $j$.

Let $x\in A$, and write $x={\sum }_i{a}_i{e}_i$ with $a_i\in F$. The dual-basis relation gives $a_j={\mathrm{Tr}}_{L\mathrm{/}F}(x{e}_j^{\ast })$ and therefore $c{a}_j={\mathrm{Tr}}_{L\mathrm{/}F}(xc{e}_j^{\ast })$. Because $x$ and $c{e}_j^{\ast }$ lie in $A$, their product is integral over $R$. The field trace of an integral element is a finite sum of its conjugates, hence is integral over $R$, and it lies in $F$; therefore [F2] forces $c{a}_j\in R$. Hence every coefficient $a_j$ lies in $c^{-1}R$.

Step 2.1 shows that $A\subseteq c^{-1}(R{e}_1+\cdots +R{e}_n)$. Multiplication by $c$ identifies this containing module with the finitely generated module $R{e}_1+\cdots +R{e}_n$, so it is finite over $R$. Since $A$ is an $R$-submodule of a finite module and $R$ is Noetherian, $A$ is finite over $R$.