Recall that a scheme $X$ is called integral at $x$ if $\mathcal{O}_{X,x}$ is an integral domain. This is equivalent to saying that $X$ is reduced at $x$ ($\mathcal{O}_{X,x}$ has no nilpotents) and that there is only one irreducible component of $X$ passing through $x$.

A scheme is called integral if it is reduced (i.e. reduced at all points) and irreducible (i.e. not a non-trivial union of two closed subsets).

It is clear that being integral is not a local property; the disjoint union of two integral schemes is not integral. This suggests the following:


Does there exist a connected scheme that is integral at all point, but not integral? (Connected means that the underlying topological space is connected.)

Note that it is easy to see that such a scheme must have infinitely many irreducible components (for else the irreducible components are open, and local integrality forces them to be disjoint).

  • 5
    $\begingroup$ This question has been asked and answered on MathOverflow:mathoverflow.net/questions/7477/… $\endgroup$ – Georges Elencwajg Sep 4 '11 at 9:52
  • $\begingroup$ For some reason I cannot manage to leave a comment (which I guess is related to the fact that I posted the question under a temporary account). Let me just say that I did try to search both mathoverflow and stackexchange; in any case the mathoverflow link is very helpful of course. $\endgroup$ – Tom Bachmann Sep 4 '11 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.