Consider the category of sheaves (of sets) on the affine étale site. It's a well known fact that a morphism of schemes is a Zariski-open immersion if and only if it is an étale monomorphism, so we extend this idea to all sheaves as follows:

We say that a sheaf $S$ satisfies the condition if given any two étale monomorphisms $A\to S$ and $B\to S$, then $A\times_S B$ is representable by $Spec(0)$ (the image of the 0 ring in the opposite category of commutative rings) if and only if either $A$ or $B$ is representable by $Spec(0)$.

Motivation: It would be nice if we could define the notion of irreducibility only in terms of functors of points. The condition that we are trying to simulate is the intersection of two open subsets of an irreducible topological space being empty if and only if one of the open subsets is empty. The problem is that the fiber-product of schemes does not necessarily coincide with the fiber product of the underlying topological space.


Are there any cases of schemes where this condition and irreducibility are not equivalent?

If this definition does work for schemes, does it work for algebraic spaces (perhaps with some tweaking)?

Edit: Recall that a morphism of sheaves $F\to G$ is an étale monomorphism if it is a monomorphism and the pullback (fiber product) by any morphism from an affine scheme $X\to G$ is an algebraic space with an atlas of affine schemes given by $\{U_i\to F \times_G X\}$ such that the composition $U_i\to F \times_G X\to X$ with the projection is an étale morphism of affine schemes (maps corresponding to étale maps of rings).

  • $\begingroup$ I have a problem with notation: what do you mean by Spec(0)? The spectrum of the 0 ring, that is, the empty scheme? And second, what do you mean when you say that fiber product of schemes does not agree with fiber product of the underlying topological spaces in the case of open immersions? Can you give a counterexample? $\endgroup$ Mar 21, 2010 at 19:10
  • $\begingroup$ I hope it's less confusing now. $\endgroup$ Mar 21, 2010 at 19:16
  • $\begingroup$ Ok, I have decided to remove all the comments, but please, try to leave things more consequential. $\endgroup$ Mar 21, 2010 at 19:36
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    $\begingroup$ Oops! An etale monomorphism, of course! (Also, in (2) I meant that such a map of sheaves is defined to be etale if it is formally smooth and locally of finite presentation.) $\endgroup$
    – JBorger
    Mar 21, 2010 at 23:55
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    $\begingroup$ OK, this is ridiculous. Apologies for all the confusion. An etale map of sheaves is defined to be one that's formally etale and locally of finite presentaiton. $\endgroup$
    – JBorger
    Mar 22, 2010 at 12:38

1 Answer 1


Yes, it is the same concept since the fibre products of schemes DO coincide with the fibre product of the underlying topological spaces in the case of open immersions (Check, for instance, Hartshornes proof for the existence of fibre products in the category of schemes).

As for the second question, I don't know how irreducibility of algebraic spaces is usually defined, or if the concept is important.

One way would be to define it as irreducibility of the underlying topological space. Look in section II.6 of Knutson to see how this is defined. He topologises the set of underlying points (equivalene classes of monomorphic K-points) by using closed immersions. Check if this is the same as topologising using open immersions (this is probably easy). If it is, my guess is that irreducibility of the underlying topological space is equivalent to your suggestion.

  • $\begingroup$ I don't see that stated anywhere in the proof. Could you give me a little more information about where it is or what implies it? $\endgroup$ Mar 22, 2010 at 10:15
  • $\begingroup$ This is a central argument in the proof; reducing to the affine case by taking affine open subschemes. It is stated and proved in step 4. $\endgroup$ Mar 22, 2010 at 10:40
  • $\begingroup$ There's more than just affine open subschemes in this case though. We're looking at all open immersions into a scheme. $\endgroup$ Mar 22, 2010 at 10:46
  • $\begingroup$ If you look at step 4, you see that it doesn't assume that anything is affine. Given a morphism of schemes f:X --> Y and an open subscheme U of Y, the pullback is given by $f^{-1}(U)$. That is, on the underlying topological spaces, it is just given by the inverse image with the relative topology, just as the topological fibre product. $\endgroup$ Mar 22, 2010 at 10:53
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    $\begingroup$ Wait, were $A$ and $B$ assumed to be schemes in the original question?! I thought they were only assumed to be sheaves on Aff in the etale topology? If they're schemes, then sure. $\endgroup$
    – JBorger
    Mar 22, 2010 at 12:41

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