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Jason Starr
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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 this$Y'''$ 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-1$$m=0,\dots,n$. The discriminant $D$ of this monic polynomial is a nonzero element of $\mathcal{O}_X(X)$. Finally, this$Y''$ is birational over $X$ to the zero scheme $Y'$ of the following section $g'$, $$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 $X$.

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 this is birational over $X$ to the zero scheme 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 $m=0,\dots,n-1$. The discriminant $D$ of this monic polynomial is a nonzero element of $\mathcal{O}_X(X)$. Finally, this is birational over $X$ to the zero scheme 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 $X$.

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 $X$.

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Jason Starr
  • 4.1k
  • 1
  • 93
  • 111

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 this is birational over $X$ to the zero scheme 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 $m=0,\dots,n-1$. The discriminant $D$ of this monic polynomial is a nonzero element of $\mathcal{O}_X(X)$. Finally, this is birational over $X$ to the zero scheme 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 $X$.

Post Made Community Wiki by Jason Starr