2
$\begingroup$

$\newcommand{\Hom}{\operatorname{Hom}}$Given a morphism of $S$-scheme, $f: X\rightarrow Y$. Consider the functor $\Hom(X,Y): ({\operatorname{Sch}}/S)^o\rightarrow (\operatorname{Sets})$ to be the following: For $S$-scheme $T,$ $\Hom(X,Y)(T)= \Hom_T(X\times_S T, Y\times_S T)$ And for morphism of $S$-schemes, $\phi: T\rightarrow T’$, $\Hom(X,Y)(\phi): \Hom_{T’}(X\times_S T’, Y\times_S T’)\rightarrow \Hom_T(X\times_S T, Y\times_S T)$, $h\mapsto h\times_{T’} \operatorname{id}_T$.

By Yoneda’s Lemma, we know that for any $S$-scheme $T$, there is a bijection between elements in $\Hom(X,Y)(T)$ and the natural transformations from $\Hom(-,T)$ to $\Hom(X,Y)$.

And Grothendieck proved that If $S$ is locally notherian, $X$ is projective and flat over $S$, $Y$ is quasi-projective over $S$, then the functor $\Hom(X,Y)$ is representable. Which means that there is a $S$-scheme $T$ such that $\Hom(-,T)=\Hom(X,Y)$. So there would be a element $\phi: \Hom_T(X\times_S T, Y\times_S T) $ such that for any $T’$ is a locally $S$-scheme, and $\psi: X\times_{S} T’\rightarrow Y\times_{S} T’$, we have unique $S$-morphism $u: T’\rightarrow T$ such that $\psi=\Hom(X,Y)(u)(\phi)$.

Then we denote $T$ by $\underline{\mathrm{Hom}}(X,Y)$. So this time we see $\underline{\mathrm{Hom}}(X,Y)$ as a scheme.

But how can we know the structure of the scheme? It mentioned on Algebraic Geometry II by Mumford that the points of $\underline{\mathrm{Hom}}(X,Y)$ are corresponding to the elements in $\Hom(X,Y)(S)$. Since it doesn’t give any details, can anyone provide an explanation for this?

$\endgroup$
  • $\begingroup$ The first thing you should do is change your notation. The last paragraph, in particular, is incomprehensible if you use the same notation $\mathrm{Hom}(X,Y)$ for a set and a functor. $\endgroup$ – Laurent Moret-Bailly Apr 17 '18 at 16:14
  • $\begingroup$ Well, I would suggest $\underline{\mathrm{Hom}}(X,Y)$ for the functor. Besides, you should not "scriptify" $X$ and $Y$ as you did in $\mathscr{Hom(X,Y)}$: this should be $\mathscr{Hom}(X,Y)$. $\endgroup$ – Laurent Moret-Bailly Apr 18 '18 at 6:43
  • $\begingroup$ @LaurentMoret-Bailly I know notation when we are talking about the Hom sheaf. That’s the reason why I said we would made some misunderstanding. Anyway, thank you for your precious advise. $\endgroup$ – Intoks Liobein Apr 18 '18 at 12:42

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.