Let $L\mid K(t_1,\ldots,t_m)$ be a separable and finite field extension with $t_1,\ldots,t_m$ algebraically independent over $K$ (algebraically closed). We can use the primitive element theorem , so there exists $t_{m+1}\in L$ such that $L=K(t_{1},\ldots,t_{m+1})$. Let $f\in K(t_{1},\ldots,t_{m})[x]$ be the minimal polynomial of $t_{m+1}$ over $K(t_{1},\ldots,t_{m})$. Then, $$ L=K(t_{1},\ldots,t_{m+1})\simeq K(t_{1},\ldots,t_{m})[x]/(f). $$ We can write $$ f=\frac{f_{0}(t_{1},\ldots,t_{m})}{g_{0}(t_{1},\ldots,t_{m})}+x\frac{f_{1}(t_{1},\ldots,t_{m})}{g_{1}(t_{1},\ldots,t_{m})}+\cdots+x^{d-1}\frac{f_{d-1}(t_{1},\ldots,t_{m})}{g_{d-1}(t_{1},\ldots,t_{m})}+x^{d}. $$ We define $h_{i}(x_{1},\ldots,x_{m})=\frac{f_{i}(x_{1},\ldots,x_{m})}{g_{i}(x_{1},\ldots,x_{m})}\cdot \operatorname{LCM}(g_{0}(x_{1},\ldots,x_{m}),\ldots, g_{d-1}(x_{1},\ldots,x_{m}))$, where LCM is the least common multiple. In particular, $h_{d}=\operatorname{LCM}(g_{0}(x_{1},\ldots,x_{m}),\ldots, g_{d-1}(x_{1},\ldots,x_{m}))$. Let $$ p(x_{1},\ldots,x_{m},x_{m+1})=h_{0}(x_{1},\ldots,x_{m})+h_{1}(x_{1},\ldots,x_{m})x_{m+1}+\cdots+h_{d}(x_{1},\ldots,x_{m})x_{m+1}^{d}\in K[x_{1},\ldots,x_{m+1}] $$ Is it true that the field of fractions of $K[x_{1},\ldots,x_{m+1}]/(p)$ is ismomorphic to $L$? Further, is it true that $(p)$ is a prime ideal?
$\textbf{Remark}$ This question has come up while I was trying to understand Theorem 5 in p.39 of Shafarevich's 'Basic Algebraic Geometry'. This theorem says that every irreducible affine set is birationally equivalent to an affine hypersurface (and this step is supposed to be obvius)
Let $R$ be a UFD, and $F$ its field of fractions. Let $F\subset L$ be a separable and finite field extension, $L=F(\alpha)$, and $f\in F[X]$ the minimal polynomial of $\alpha$ over $F$. Then let $a\in R$ be the least common multiple of the denominators of coefficients of $f$, and set $p=af\in R[X]$.
Is $(p)$ a prime ideal? Yes, it is: $p$ is irreducible in $F[X]$ and primitive.
The field of fractions of $R[X]/(p)$ is isomorphic to $L$? Yes, it is. (See this answer.)