Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $X$ is an object in an arbitrary category, there is a natural definition of a subobject of $X$ as a monomorphism into $X$ (or really an equivalence class of monomorphisms). If $X$ is a scheme, however, the term 'subscheme' is conventionally reserved only for locally closed immersions (as in EGA I.4.1.2). There are certainly many monomorphisms of schemes that in this sense aren't subschemes, for example the inclusion of Spec of a local ring such as $Spec K[x]_{(x)} \to Spec K[x]$.

When we restrict 'subscheme' to mean 'locally closed immersion', defining images of schemes becomes problematic. A sensible definition, in any category, of the image of a morphism is the minimal subobject through which it factors. Using the above definition of subscheme, there are perfectly well-behaved examples of morphisms of schemes that don't have images in this sense. For example, consider the morphism $\mathbb A^2_K \to \mathbb A^2_K $ induced by the ring homomorphism $(x,y) \mapsto (x,xy)$; the set-theoretic image is the union of the origin and the complement of the $y$-axis, and there is no minimal locally closed set containing this.

There is, however, always a minimal closed immersion through which a given morphism factors, and so if one defines the scheme-theoretic image in this sense, it always exists. My question is that, if we let our notion of 'subscheme' include all monomorphisms, would the resulting notion of 'scheme-theoretic image' always exist? In other words, is there always a minimal monomorphism of schemes through which a given morphism factors? Say, in the above example? If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?

As a 'softer' question, can someone explain why we don't want to call general monomorphisms subschemes? In particular, suppose I have a morphism that is a submersion onto a locally-but-not-globally closed subscheme. It seems much more sensible to call that locally closed subscheme the image, rather than its global closure.

share|improve this question
"If I hand you a constructible subset of a scheme, can you only find a monomorphism onto that set if it's locally closed?". I'm not so sure. Let me try this example: Let X be affine 2-space, let U be X minus the y-axis and let P be the origin. Then U union P is a nice example of a constructible set: it's the image of the map U disjoint P -->X (open immersion on U, closed immersion on P) and it's not locally closed. But I think the map (U disjoint union P) --> X is a monomorphism; if I have two maps T-->(U disjoint P) which become equal on X then write T=T_U disjoint T_P and follow your nose. –  Kevin Buzzard Mar 30 '10 at 20:23
re subject of the previous comment: for any constructible (indeed even pro- or ind-constructible) subset of a scheme X there is a monomorphism of schemes W->X whose image is that set. the reason is (see EGAIV.I.1.9, or 1971 EGAI.7.2) that for any scheme X there exists a scheme X^cons, a monomorphism X^cons -> X which is bijective on points and which has the property that the map induced on point sets gives a bijection between closed (resp. open, resp. open and closed) subsets of X^cons and pro-constructible (resp. ind-constructible, resp. constructible) subsets of X. –  LRG Mar 30 '10 at 22:55
A definition of "image" whose formation is not visibly local on the target (i.e., formation does not visibly commute with base change alone open immersions) is unlikely to be of much use in proofs, let alone to do anything interesting. This is but one of the defects of the definition in terms of global minimal factorization through a closed immersion. Likewise, the reason not all monomorphisms are called subschemes is because a general monomorphism is a crazy thing. Remember that the subject is called algebraic geometry, not algebraic category theory. –  BCnrd Mar 31 '10 at 3:08
add comment

1 Answer

As a partial answer to your "softer" question, general monomorphisms (as Brian Conrad points out) can be more general than we really want subschemes to be. For instance, if $A$ is a Noetherian ring and $\hat{A}$ its completion with respect to some ideal, then $\hat{A}$ is isomorphic to $\hat{A} \otimes_A \hat{A}$. Consequently, Spec $\hat{A}$ is isomorphic to $\mathrm{Spec} \hat{A} \times_{\mathrm{Spec} A} \mathrm{Spec} \hat{A}$, and so Spec $\hat{A} \to $ Spec $A$ is a monomorphism of schemes. However, I do not think we want to call Spec $\hat{A}$ a subscheme of Spec $A$.

Edit: As the comment below points out, this example is incorrect. I had assumed that $\hat{A} \otimes_A \hat{A} = \hat{\hat{A}} = \hat{A}$, but the theorem I was using would require that $\hat{A}$ be a finitely generated $A$-module, which is typically not the case. However, the same reasoning shows that, for instance, Spec $k(t) \to$ Spec $k[t]$ is a monomorphism, whereas one would expect that any subscheme of a variety should contain at least one closed point.

share|improve this answer
Careful, $\widehat{A} \otimes_ A \widehat{A}$ is generally much much much bigger than $\widehat{A}$, and typically not even noetherian (and ${\rm{Spec}} \widehat{A} \rightarrow {\rm{Spec}} A$ is generally very far from a monomorphism, with gigantic geometric fibers). Due to its importance in fpqc descent, this was one of the key examples which motivated Grothendieck to change his mind about the necessity of noetherian hypotheses (as in his 1958 ICM talk). On the other hand, a localization at a multiplicative set provides a typical example of the sort I think you wanted to express. –  BCnrd Apr 2 '10 at 7:49
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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