# Coarse moduli spaces and rational points [closed]

Let $K$ be a field (not necessarily algraically closed). Let $\mathcal{F}$ be a contravariant functor from the category of schemes over $K$ to sets and $M$ be a coase moduli space for the functor. So, there is a morphism of contravariant functors from $\mathcal{F}$ to $h_M$ where $h_M:=Hom(−,M)$ such that $\mathcal{F}(\bar{K})$ is bijective to $h_M(\bar{K})$, where $\bar{K}$ is an algebraic closure of $K$. So, the natural morphism from $\mbox{Spec}(\bar{K})$ to $\mbox{Spec}(K)$ gives rise to a natural transformation. As far as I understand this means that if $\mathcal{F}(K)$ is non-empty which maps to something non-empty in $\mathcal{F}(\bar{K})$ the image of $\mathcal{F}(K)$ in $h _M(\bar{K})$ under the composition of the morphism $\mathcal{F}(K) \to \mathcal{F}(\bar{K}) \xrightarrow{\sim} h_M(\bar{K})$ is non-empty (due to the bijectivity over $\bar{K}$). By the commutativity of the diagram (coming from the natural transformation) this means that $h_M(K)$ is non-empty. Is this correct?

-

## closed as off-topic by Neil Strickland, Daniel Loughran, Steven Sam, Stefan Kohl, abxJun 8 '14 at 21:23

• This question does not appear to be about research level mathematics within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

Can't you just say that you have a map $\mathcal{F}(K)\rightarrow h_M(K)$, so $h_M(K)$ is non-empty if $\mathcal{F}(K)$ is non-empty? –  abx Jun 8 '14 at 8:06
Indeed, that was the answer given at the duplicate on MSE. –  Zhen Lin Jun 8 '14 at 8:31
@Lin, abx: However, I do not entirely understand the answer. My confusion is that $h_M(K)$ contains the empty set. So, why cant $\mathcal{F}(K)$ be that? –  user45397 Jun 8 '14 at 10:36