A finite-type field reduces to a localization over a transcendence basis

Choose generators $a_1,\dots ,a_m\in K$ of $K$ as a $k$-algebra. Since $t_1,\dots ,t_r$ is a transcendence basis, [L2] shows that each $a_i$ is algebraic over $F:=k(t_1,\dots ,t_r)$.

Put $R=k[t_1,\dots ,t_r]\subseteq K$. The ratios $b\mathrm{/}c$ with $b,c\in R$ and $c\mathrm{\ne }0$ form a subfield of $K$ containing $R$; conversely every such ratio lies in $F$, so this subfield equals $F$. For each of the finitely many $a_i$, clear the finitely many denominators in an algebraic equation over $F$. This gives a nonzero polynomial $c_{i,d_i}{X}^{d_i}+\cdots +c_{i,0}\in R[X]$ vanishing at $a_i$, with $d_i\ge 1$ and $c_{i,d_i}\mathrm{\ne }0$. Put $s={\prod }_i{c}_{i,d_i}$, taking $s=1$ for an empty generating list. Since $R$ is a subring of the field $K$, $s\mathrm{\ne }0$.

By [L4], $A=R[1\mathrm{/}s]$ embeds in $K$ as the fractions $b\mathrm{/}{s}^j$: the extension map sends $b\mathrm{/}{s}^j$ to the same quotient in $K$, and a zero image forces $b=0$. Each leading coefficient has inverse $({\prod }_{h\mathrm{\ne }i}{c}_{h,d_h})\mathrm{/}s$ in $A$. Dividing its equation by that coefficient makes it monic, so every $a_i$ is integral over the nonzero ring $A$ by [L3].

Every $a\in A$ is integral over $A$ by the polynomial $X-a$. Thus the integral subring of $K$ from [L3] contains $A$ and all the $a_i$, hence contains $A[a_1,\dots ,a_m]$. Since $k\subseteq A\subseteq K$ and $k[a_1,\dots ,a_m]=K$, this generated algebra equals $K$. Therefore every element of $K$ is integral over $A$.

Thus $K$ is integral over $k[t_1,\dots ,t_r][1\mathrm{/}s]$.