Let $X$ be an integral Noetherian separated scheme. Under what conditions can we find a nonempty affine open in $X$ whose complement is irreducible?
1 Answer
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 $XH$ is the nonempty, 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 quasiprojective then the same argument applies, after passing to a projective closure $\bar{X}$ of $X$.

$\begingroup$ here "projective" means "projective over an algebraically closed field", right? $\endgroup$– user138661Apr 26, 2019 at 12:55

$\begingroup$ 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/… $\endgroup$ Apr 26, 2019 at 13:01

$\begingroup$ 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. $\endgroup$– user138661Apr 26, 2019 at 13:05

$\begingroup$ "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 nonalgebraically closed fields, $X$ must be geometrically irreducible. $\endgroup$ Apr 26, 2019 at 13:15