# Affine open with irreducible complement

Let $$X$$ be an integral Noetherian separated scheme. Under what conditions can we find a non-empty affine open in $$X$$ whose complement is irreducible?

An (almost) obvious sufficient condition is that $$X$$ is projective.

In fact, take a very ample line bundle $$\mathcal{H}$$ on $$X$$. Then the general element $$H$$ in the complete linear system $$|\mathcal{H}|$$ is irreducible because, by a version of Bertini theorem, a reducible linear system without fixed parts is necessarily composed with a pencil, in particular, it cannot be very ample (see [O. Zariski, Algebraic surfaces, page 26]).

Now $$X-H$$ is the non-empty, open affine subset you are looking for.

Remark 1. The argument works for geometrically irreducible projective schemes over any field (included finite fields, see the comments).

Remark 2. If $$X$$ is quasi-projective then the same argument applies, after passing to a projective closure $$\bar{X}$$ of $$X$$.

• here "projective" means "projective over an algebraically closed field", right?
– user138661
Apr 26, 2019 at 12:55
• There are versions of Bertini's theorem working even over finite fields (under the assumption that $X$ is geometrically irreducible, i.e., irreducible over any field extension of the base field). See pdfs.semanticscholar.org/87ee/… Apr 26, 2019 at 13:01
• but I mean the question does not specify the base (the scheme could be even $\mathrm{Spec}\,\mathbb{Z}$) so you probably could specifically say that in the answer.
– user138661
Apr 26, 2019 at 13:05
• "Over a field" is enough. Classical Bertini works for infinite fields, and the case of finite fields is addressed by the link I posted in the comment above. Over non-algebraically closed fields, $X$ must be geometrically irreducible. Apr 26, 2019 at 13:15