3
$\begingroup$

T am trying to refram what I have learned in algebraic geometry in the context of the functor of points approach. In particular, I want to practice proving the existence of a scheme satisfying a certain universal property by showing a corresponding functor is representable. The most common example of this in textbooks is the fibered product of schemes. If $X \rightarrow S$ and $Y \rightarrow S$ are schemes over $S$, then to show that the fibered product over $S$ exists is to show that the functor $$ \hom(-, X) \times_{\hom(-, S)} \hom(-Y) $$ is representable. This is done by showing that the functor is covered by open representable subfunctors, and that the functor is a sheaf in the Zariski topology. The representing object is then the desired scheme.

My problem is that this method doesn't seem to be very practical in the vast majority of cases. My understanding was that the functor of points approach was preferred in modern abstract algebraic geometry.

The problem I am currently trying to do is to prove that the normalization of a scheme exists via this method. For the sake of keeping things simple, let's just look at integral schemes, and we will define "normal" to mean all the local rings are integrally closed domains.

So let's say that $X$ is an integral scheme. I want to find some functor $$ F: \text{Sch}^{\text{op}} \longrightarrow \text{Set} $$ which captures the universal property.

The universal property in this case will be something like "a scheme $\tilde{X}$ over $X$ is called the normalization of $X$ if every dominant morphism morphism $Y \rightarrow X$ with $Y$ normal factors uniquely through $\tilde{X}$".

My question is, how do I turn that into a representable functor? Clearly it must be possible, since every scheme can be realized that way. Is this just the wrong situation to use it in to make life easier? The universal property isn't even in terms of schemes over $X$, but in terms of very specific types of schemes with dominant structure morphisms. So it seems any representability criterion will need to be modified. Is this even a sensible thing to be trying to do?

$\endgroup$
  • $\begingroup$ Here is an alternate question for which again I do not see such a functor. Take a smooth curve $C$ (over an algebraically closed field, $k$). Is there a functor from $\mathrm{Sch}^{op}$ to sets which will capture the unique projective completion of $C$? May be I am missing something. $\endgroup$ – Mohan Dec 19 '18 at 2:39
  • $\begingroup$ It's an interesting question. You might try it at MO. $\endgroup$ – Kevin Carlson Dec 20 '18 at 19:47

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.