# Algebraic spaces as functors on complete local rings

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?

• How do you define the category of complete local rings? What are the arrows? – Angelo Sep 12 at 16:56
• Ordinary homomorphisms of rings, not necessarily local homomorphisms. – Mellon Sep 12 at 17:10
• 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. – Anonymous Sep 12 at 18:29
• 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. – Mellon Sep 12 at 19:04

• What happens at the generic point of $\mathbb{A}^1$? The fraction field is a complete local ring. Or did I misunderstand the question? – Piotr Achinger Sep 14 at 12:44
• 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 $X$. – Jason Starr Sep 15 at 0:03