Images and Monomorphisms of Schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:59:05Z http://mathoverflow.net/feeds/question/19871 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19871/images-and-monomorphisms-of-schemes Images and Monomorphisms of Schemes Harold Williams 2010-03-30T19:38:04Z 2010-04-02T14:28:12Z <p>If $X$ is an object in an arbitrary category, there is a natural definition of a <a href="http://en.wikipedia.org/wiki/Subobject" rel="nofollow">subobject</a> 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 <code>$Spec K[x]_{(x)} \to Spec K[x]$</code>.</p> <p>When we restrict 'subscheme' to mean 'locally closed immersion', defining images of schemes becomes problematic. A sensible definition, in any category, of the <a href="http://en.wikipedia.org/wiki/Image_%28category_theory%29" rel="nofollow">image</a> 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 <code>$\mathbb A^2_K \to \mathbb A^2_K$</code> induced by the ring homomorphism <code>$(x,y) \mapsto (x,xy)$</code>; 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. </p> <p>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, <a href="http://math.columbia.edu/algebraic_geometry/stacks-git/morphisms.pdf#01R6" rel="nofollow">it always exists</a>. 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?</p> <p>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. </p> http://mathoverflow.net/questions/19871/images-and-monomorphisms-of-schemes/20140#20140 Answer by Charles Staats for Images and Monomorphisms of Schemes Charles Staats 2010-04-02T04:36:57Z 2010-04-02T14:28:12Z <p>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 $A$.</p> <p>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.</p>