Let $X \to S$ be a morphism of schemes and define the following functor $$ F: (\operatorname{Aff}/X)^{\operatorname{opp}}\to \operatorname{Set}$$ $$ (f:\operatorname{Spec}A \to X) \mapsto \{ u:\operatorname{Spec}A[\epsilon] \to X | u \circ \pi=f\}$$ where $\pi$ is the associated map to the canonical projection $A[\epsilon] \to A$. I want to show that this functor is representable by the relative spectra of the sheaf of $\mathcal{O}_X$-algebras $ \operatorname{Sym}\Omega^1_{X/S}$. This is pretty straightforward if we assume $X,S$ to be affine.

I would like to find a way to reduce the problem to this simpler case. To do this I fix a covering $\{U_i\}_{i \in I}$ of $X$ by affine opens such that the image of every $U_i$ lies in some affine open of $S$. Then I define the following family of open subfunctors of $F$ via, $$F_i= F \times_{h_X} h_{U_i}$$ where $h_X= \operatorname{Hom}_{ \operatorname{Sch/U}}(-,U)$. Let us remark that $h_X$ is terminal in the category of schemes over $X$ and that we have a map induced by the inclusion of $U_i$ into $X$ namely $h_{U_i} \to h_X$. Given $f:\operatorname{Spec}A \to X$ then we have that $$F_i(f:\operatorname{Spec}A \to X)=\emptyset \text{ if f does not factor through } U_i$$ and $$F_i(f:\operatorname{Spec}A \to X)=F(f:\operatorname{Spec}A \to X) \text{ otherwise.}$$ If $f:\operatorname{Spec}A \to X$ factors through $U_i$ then if I show that every $u \in F(f:\operatorname{Spec}A \to X)$ must factor through $U_i$ then I can reduce it to the affine case. However I don't see how I could show this factorization despite the fact that the topological spaces of $\operatorname{Spec}A ,\operatorname{Spec}A[\epsilon]$ are homeomorphic.

Can one show this? Is there any other way of producing a good reduction to the affine case?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.