Let $X$ be a geometrically irreducible, possibly singular projective variety over an infinite field $k$. Assume that the dimension of $X$ is at least 2. Can there exist a hyperplane section of $X$ that is not geometrically connected?
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1$\begingroup$ What is "geometrically connected"? $\endgroup$– Zach TeitlerCommented Feb 28, 2020 at 20:29
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3$\begingroup$ Just take a smooth quadric $Q\subset \mathbb{P}^3$, a point $p\in Q$, $X=Q\smallsetminus\{p\}$, and the tangent hyperplane to $Q$ at $p$. $\endgroup$– abxCommented Feb 28, 2020 at 20:36
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$\begingroup$ @abx But $Q \setminus \{p\}$ is not a projective variety? $\endgroup$– Kevin CastoCommented Feb 28, 2020 at 23:30
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$\begingroup$ @KevinCasto the question was edited in response to abx's comment. $\endgroup$– user145520Commented Feb 28, 2020 at 23:31
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$\begingroup$ Anyways, you may be able to craft an answer from part A of mathoverflow.net/questions/114898/… $\endgroup$– Kevin CastoCommented Feb 29, 2020 at 3:32
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1 Answer
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Kevin Casto's comment appears to do the trick. But in case a reference would be helpful, here is a result from this paper by Martinelli, Naranjo, and Pirola:
Theorem 1.1: Let $k$ be an algebraically closed field, $X$ an irreducible projective variety over $k$, and $f:X\to\mathbb{P}_k^n$ a morphism. If $r\geq n+1-\text{dim}\,f(X)$, then for any $r$-dimensional linear subvariety $L\subseteq \mathbb{P}_k^n$, $f^{-1}(L)$ is connected.
In particular, if $\text{dim}\,f(X)\geq 2$, every hyperplane section is geometrically connected.
