A finite-type domain over a field has finite normalization

By [L5] the finite-type domain $A$ over $k$ is nonzero, so [L1] applies: fix algebraically independent elements $z_1,\dots ,z_d\in A$ such that $A$ is module-finite over $R:=k[z_1,\dots ,z_d]$. Every $a\in A$ is integral over $R$: the ring $A$ is a faithful $R[a]$-module, because $r\cdot 1_A=r\mathrm{\ne }0$ for every nonzero $r\in R[a]\subseteq A$ (evaluate at $1_A$ in the domain $A$), and it is a finite $R$-module, so [L2] applies to $b=a$ inside $R\subseteq A$. Hence $R\subseteq A\subseteq \mathrm{Frac}(A)$ with $A$ integral over $R$.

The extension $F:=\mathrm{Frac}(R)$ is a rational function field and $\mathrm{Frac}(A)\mathrm{/}F$ is finite. Write $A=R{a}_1+\cdots +R{a}_n$ with $a_i\in A$, using the module finiteness of step 1.1. Every element of $A$ lies in $F[a_1,\dots ,a_n]\subseteq F(a_1,\dots ,a_n)$, and $F(a_1,\dots ,a_n)$ is a field containing $A$, so $\mathrm{Frac}(A)=F(a_1,\dots ,a_n)$ by [L5]; each $a_i$ is integral over $R$ by step 1.1, hence algebraic over $F$; therefore $\mathrm{Frac}(A)\mathrm{/}F$ is finite by [L6].

Apply [L3] with the base field $k$, the algebraically independent elements $z_1,\dots ,z_d$ (so that $R=k[z_1,\dots ,z_d]$ and $F=k(z_1,\dots ,z_d)$) and the finite extension $L:=\mathrm{Frac}(A)$ of step 2.1: the integral closure $B$ of $R$ in $\mathrm{Frac}(A)$ is a finite $R$-module.

Since $A$ is integral over $R$ by step 1.1 and $R\subseteq A\subseteq \mathrm{Frac}(A)$, [L4] shows that an element of $\mathrm{Frac}(A)$ is integral over $R$ exactly when it is integral over $A$; hence $B$ is exactly the integral closure of $A$ in $\mathrm{Frac}(A)$, and $B$ is a ring with $R\subseteq A\subseteq B$. If $B=R{b}_1+\cdots +R{b}_r$, then every $A{b}_i$ lies in $B$, and every $R{b}_i$ lies in $A{b}_i$ because $R\subseteq A$; hence $B=A{b}_1+\cdots +A{b}_r$ is generated as an $A$-module by the same finitely many elements. Therefore the integral closure of $A$ in $\mathrm{Frac}(A)$ is a finite $A$-module.