Does every finite map of smooth varieties birationally embed as a smooth hypersurface of 1d trivial bundle over the base?

Question

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 $\mathbb C(X) \subset \mathbb C(Y)$?

Answer 1

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 $X$.

Answer 2

I may have misunderstood something. However my impression is that, since you are asking $Y'$ smooth, the answer is negative.

The simplest counterexamples are probably compact Riemann surfaces, since birational maps among them are isomorphisms. In fact if I'm not wrong, for $X={\mathbb P}^1$, this happens only if the finite morphism $Y \to X$ is an isomorphism.

Here is a proof. Since $X={\mathbb P}^1$ then $Y' \subset {\mathbb P}^1 \times {\mathbb A}^1 \subset {\mathbb P}^1 \times {\mathbb P}^1$ is an irreducible divisor that never intersects the section at $\infty$. But the only irreducible divisors in ${\mathbb P}^1 \times {\mathbb P}^1$ with this property are the sections, so $Y'$ must be isomorphic to $X={\mathbb P}^1$.