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Let $k$ be a field and let $V$ a finite-dimensional vector space over $k$. Assume that $X\subset \mathbb{P}(V)$ is a closed subvariety. Does there exist a proper flat morphism $Y\to \mathbb{P}(V^*)$ such that the fiber over a closed point $p\in \mathbb{P}(V^*)$ corresponding to a given hyperplane in $\mathbb{P}(V)$ is isomorphic to the hyperplane section of $X$ cut out by that hyperplane?

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    $\begingroup$ Flatness won't hold, since the dimension of the fibers will probably change. Have you taken a look at (arxiv.org/pdf/math/0504020.pdf)? $\endgroup$
    – Eoin
    Commented Mar 2, 2020 at 0:52

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Let $I \subset \mathbb{P}(V) \times \mathbb{P}(V^\vee)$ be the incidence divisor (a smooth divisor of bidegree $(1,1)$). Then $$ Y = X \times_{\mathbb{P}(V)} I $$ is the universal family of hyperplane sections of $X$. It is flat over $\mathbb{P}(V^\vee)$ if and only if irreducible components of $X$ do not lie in hyperplanes of $\mathbb{P}(V)$.

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