3
$\begingroup$

Let $X$ be an algebraic space locally of finite presentation, and let $\tilde{X}$ denote the restriction of $X$ (as a functor on schemes) to the category of complete local rings. Is it true that the mapping $X \mapsto \tilde{X}$ (of algebraic spaces to functors on complete local rings) is a fully faithful functor?

I.e. can we uniquely determine a morphism $f : X \to Y$ of algebraic spaces locally of finite presentation simply by specifying its values on complete local rings?

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ How do you define the category of complete local rings? What are the arrows? $\endgroup$
    – Angelo
    Commented Sep 12, 2020 at 16:56
  • $\begingroup$ Ordinary homomorphisms of rings, not necessarily local homomorphisms. $\endgroup$
    – Mellon
    Commented Sep 12, 2020 at 17:10
  • 6
    $\begingroup$ This cannot be true without any finiteness conditions. For example, take a non-reduced ring $R$ with a unique prime $m$ satisfying $m=m^2$ (e.g, a quotient of a rank $1$ nondiscrete valuation ring by a nonzero nonunit). Then any map $R \to S$ to a local ring $(S,n)$ is necessarily a local map, and thus kills $m$ if $S$ is also complete (as $m$ maps into $n^k$ for all $k$ since $m=m^k$). So the spectrum of both $R$ and $R/m$ represent the same functor on complete local rings. $\endgroup$
    – Anonymous
    Commented Sep 12, 2020 at 18:29
  • $\begingroup$ I suppose the Noetherian is assumption will eliminate these types of counter-examples? I will modify the question, there should reasonably be at least some finiteness assumptions. $\endgroup$
    – Mellon
    Commented Sep 12, 2020 at 19:04

1 Answer 1

Reset to default
6
$\begingroup$

There is an exponential map from complex line to itself. This is not algebraic, but it is defined on complete local ring valued points. So the functor does not appear to be full.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ What happens at the generic point of $\mathbb{A}^1$? The fraction field is a complete local ring. Or did I misunderstand the question? $\endgroup$
    – Piotr Achinger
    Commented Sep 14, 2020 at 12:44
  • $\begingroup$ I agree with Achinger. My answer works only for residue field being finite over the base. $\endgroup$
    – Nitin Nitsure
    Commented Sep 14, 2020 at 12:56
  • $\begingroup$ You can make an example where $X$ is a nodal plane cubic and where $Y$ is the quotient of the nontrivial double cover by the involution acting only on the smooth locus. Thus, the algebraic space $Y$ is nonseparated with two closed points over the node of $X$. There is a “section on complete local rings” of the natural morphism from $Y$ to $X$. $\endgroup$
    – Jason Starr
    Commented Sep 15, 2020 at 0:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .