# Bertini theorem for connectedness

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?

• What is "geometrically connected"? Commented Feb 28, 2020 at 20:29
• 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 $p$.
– abx
Commented Feb 28, 2020 at 20:36
• @abx But $Q \setminus \{p\}$ is not a projective variety? Commented Feb 28, 2020 at 23:30
• @KevinCasto the question was edited in response to abx's comment.
– user145520
Commented Feb 28, 2020 at 23:31
• Anyways, you may be able to craft an answer from part A of mathoverflow.net/questions/114898/… Commented Feb 29, 2020 at 3:32

## 1 Answer

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.