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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?

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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 $X$.

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  • $\begingroup$ here "projective" means "projective over an algebraically closed field", right? $\endgroup$
    – user138661
    Apr 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. $\endgroup$
    – user138661
    Apr 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 non-algebraically closed fields, $X$ must be geometrically irreducible. $\endgroup$ Apr 26, 2019 at 13:15

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