Suppose $Y \to X$ is a finite morphism of varieties over $\mathbb C$, with $Y$ and $X$ both smooth. Is $Y$ always birational to a smooth hypersurface $Y' \subset X \times \mathbb A^1$, such that the projection $Y' \to X$ induces $\mathbb C(X) \subset \mathbb C(Y)$?



Does every finite map of smooth varieties birationally embed as a smooth hypersurface of 1d trivial bundle over the base?
This is an answer to the question in the original post. The answer is positive in many cases, in particular whenever $X$ is affine. There is nothing to prove if the degree $n$ of the finite morphism equals $1$. Thus, assume that the degree $n$ is at least $2$.


By the Primitive Element Theorem, $Y$ is birational over $X$ to a hypersurface $Y'''$ in $X\times \mathbb{A}^1$ with its first projection to $X$, i.e., $Y'''$ is the zero scheme of a global section $g'''$ of the structure sheaf $$g\_n(a\_n,\dots,a\_m,\dots,a\_1,a\_0;t) = a\_nt^n + \dots + a\_mt^m + \dots + a\_1t+a\_0$$ for $(a\_n,\dots,a\_m,\dots,a\_1,a\_0) \in \mathcal{O}\_X(X)^{\oplus(n+1)}$ with $a\_n$ and $a\_0$ both nonzero, and where $\mathbb{A}^1=\text{Spec}\ \mathbb{Z}[t]$. Of course $Y'''$ is birational over $X$ to the zero scheme $Y''$ of the following *monic* section $g''$, $$g\_n(1,\dots,b\_m,\dots,b\_1,b\_0;t) = a\_n^{n-1}g\_n(a\_n,\dots,a\_m,\dots,a\_1,a\_0;t/a\_n) = $$ $$t^n + \dots + b\_mt^m + \dots + b\_1t+b\_0,$$ where $b\_m$ equals $a\_ma\_n^{n-m-1}$ for $m=0,\dots,n$. The discriminant $D$ of this monic polynomial is a nonzero element of $\mathcal{O}\_X(X)$. Finally, $Y''$ is birational over $X$ to the zero scheme $Y'$ of the following section $g'$, $$g\_n(D^n,\dots,c\_m,\dots,c\_1,c\_0;t) = D^ng\_n(1,\dots,b\_m,\dots,b\_1,b\_0;t+(1/D)),$$ where $c\_m$ is the section $$c\_m=\sum\_{\ell=0}^{n-m}\binom{m+\ell}{\ell}D^{n-\ell}b\_{m+\ell}.$$ By construction, the zero scheme $Y'$ of $g'$ is étale over $X$.


