Consider a morphism of schemes $f \colon X \to Y$ having the following properties:
i) Topologically, $f$ is a homeomorphism onto its image.
ii) The induced map of sheaves $\mathcal{O}_Y \to f_* \mathcal{O}_X$ $f^{-1}\mathcal{O}_Y \to \mathcal{O}_X$ is surjective. (See Wedhorn's answer.)
iii) $f$ is [edit: locally] of finite type. (This excludes morphisms such as $\mathrm{Spec} \; K(X) \to X$ with image the generic point of the variety $X$.)

Clearly, any immersion satisfies these three properties.

I have two questions:
1) Is $f$ necessarily an immersion (i.e., closed immersion followed by an open immersion)?
2) If not, are there important properties satisfied by immersions, but not necessarily by morphisms satisfying the three properties above?

Motivation: I find the three properties above more "natural" to work with than the property of being a closed immersion followed by an open immersion. For instance, it is immediately obvious that the composition of two morphisms satisfying i)-iii) also satisfies i)-iii); I do not find this immediately obvious for immersions in the usual sense (although I will be pleased, if not entirely surprised, if some respondents make this obvious).

  • 2
    $\begingroup$ Your condition ii) is rather strange: it is satisfied for closed embedding, obviously, but not for open ones, in general. $\endgroup$
    – Angelo
    Mar 19, 2011 at 16:12
  • 1
    $\begingroup$ Perhaps (ii) should be replaced with: All stalk maps $\mathcal{O}_{Y,f(x)} \to \mathcal{O}_{X,x}$ are surjective. $\endgroup$ Mar 19, 2011 at 16:39
  • 1
    $\begingroup$ As for compositions of immersions being immersions, that is easy to see. All you need is to show that any f, which is an open immersion followed by a closed immersion can be factored as a closed immersion followed by an open immersion. The image of your open immersion is an open subset of im f and therefore comes as an intersection of an open set U with im f. Well, in your new factorization U will be give an open immersion and im f \cap U will give a closed immersion (into U). $\endgroup$ Mar 19, 2011 at 16:49
  • $\begingroup$ It is not true that immersions can always be factorized into an open immersion followed by a closed immersion. A nice example, where this is not possible, can be found in the Stacks project (www.math.columbia.edu/algebraic_geometry/stacks-git) Example 22.2.10. $\endgroup$ Mar 19, 2011 at 17:28

1 Answer 1


As Angelo already pointed out, (ii) is not satisfied by open immersions (e.g. take $R$ a dvr with field of fractions $K$ and let ${\rm Spec} K \to {\rm Spec} R$ be the canonical immersion). In addition not all open immersions are quasi-compact. In particular, they are not necessarily of finite type. Here counterexamples are more difficult to write down as this cannot happen if $Y$ is noetherian.

But nevertheless you were almost there: Immersions $f\colon X \to Y$ can be characterized by the following two properties:

(i) $f$ induces a homeomorphism from $X$ onto a locally closed subspace of $Y$.

(ii) $f^{-1}{\mathcal O}_Y \to {\mathcal O}_X$ is surjective.

This is shown in EGA I, (4.2.2).

  • 1
    $\begingroup$ If we eliminate the "locally closed" requirement, what utility do we lose in the resulting notion? $\endgroup$ Mar 19, 2011 at 17:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.