# For a hyperplane section Z of X, when there exists its etale X-neighbourhood such that Z is a fibre of its morphism to a curve

Let $X$ be a smooth projective variety (over an algebraically closed field; it could be the field of complex numbers); $Z$ is its hyperplane section. When there exists an etale $U/X$ such that:

1. $Z$ can be lifted to $U$.

2. There exists a morphism from $U$ to a curve $C$ such that $Z$ is the preimage of some point of $C$.

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I am not sure I understand the question. If you had such an etale neighborhood of $Z$, then the normal bundle of $Z$ in $U$ will be trivial. But the normal bundle to $Z$ in $U$ is isomorphic to the normal bundle of $Z$ in $X$, and $N_{Z/X}$ can never be trivial for a hyperplane section. –  Tony Pantev Sep 22 '12 at 14:51

If $X$ is a curve, then just take linear projection from a disjoint codimension 2 linear subspace in the spanning hyperplane of $Z$. If the dimension of $X$ is $2$ or more, then there never exists such an étale neighborhood and morphism. For $Z$ of dimension $n-1 \geq 1$, for the normal sheaf $\mathcal{N}_{Z/X}$, the top intersection number $c_1(\mathcal{N}_{Z/X} )^{n-1}$ equals the degree of $X$, which is positive. Since $\mathcal{N}_{Z/X}$ equals $\mathcal{N}_{Z/U}$, also $c_1(\mathcal{N}_{Z/U})^{n-1}$ is positive. However, for a fiber of a morphism, $\mathcal{N}_{Z/U}$ is isomorphic to $\mathcal{O}_Z$ so that $c_1(\mathcal{N}_{Z/U})$ equals $0$.
Suppose that $H$ is an hyperplane such that $Z = H \cap X$. Let $L$ be a subspace of codimension 1 of $H$; the pencil of hyperplanes passing through $L$ defines a morphism $X \smallsetminus L \to \mathbb P^1$ such that the inverse image of a point is $Z \smallsetminus L$. If you want $U$ to surject onto $X$ you can cover $X$ with various $X \smallsetminus L$, and take the disjoint union.
: I misunderstood, the OP wants $Z$ to be embedded in $U$. This implies that the normal bundle of $Z$ in $X$ is trivial, and this never happens, unless $X$ is a curve.
No, I don't need a surjection of $U$ onto $X$. On the other hand, I need a closed embedding $Z\to U$; $Z\setminus L\subset U$ is not enough. –  Mikhail Bondarko Sep 22 '12 at 14:25