Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ and $Y$ be projective schemes. Then we can define the mapping scheme between them, $\rm{Maps}(X,Y)$ as follows:

To any map $f:X\rightarrow Y$ we consider the graph $\Gamma_f$ as a closed subscheme of $X \times Y$. So $\rm{Maps}(X,Y)$ is the set of all subschemes of $X \times Y$ that are graphs of morphisms. (Concretely, a subscheme $Z \subset X \times Y$ is the graph of a morphism iff the projection to $X$ is an isomorphism) Of course this all makes sense in families, so $\rm{Maps}(X,Y)$ is a subfunctor of the Hilbert scheme $\rm{Hilb}(X \times Y)$.

Now at this point, I have seen a number of sources casually claim that $\rm{Maps}(X,Y)$ is actually an $\it{open}$ subfunctor and is hence representable. None of these sources even remark on why this is true? So my question is: why is this true?

share|improve this question
    
Which Hilbert scheme do you (and Kollar) mean exactly? I am new to this and only know Hilbert schemes w.r.t. some polynomial... Would you take the Hilbert polynomial of the ideal sheaf of some subspace $X \times \{pt\}$ (assuming that $Y$ has a point)? –  Peter Arndt Sep 29 '10 at 20:17
    
For the above statement you can just take the union of every Hilbert schemes, i.e. of the Hilbert scheme for every possible Hilbert polynomial. (Similarly, Maps(X, Y) contains all maps, not just maps of some fixed degree.) –  Arend Bayer Sep 29 '10 at 22:03

2 Answers 2

See Grothendieck, seminaire bourbaki 221 "les schemas de Hilbert", bottom of page 221-19.

share|improve this answer
2  
Grothendieck is a bit sketchy there. A more detailed argument fleshing out his sketch is given in Theorem 5.23 in the book FGA Explained (for which the preceding result 5.22(b) provides the crucial openness result). –  BCnrd Sep 30 '10 at 0:43
3  
First time I've seen anyone imply that Grothendieck doesnt write enough. –  Richard Borcherds Sep 30 '10 at 1:39
2  
Dear Richard: I think the FGA's weren't meant to be more than sketches (unlike EGA). I didn't intend to suggest he didn't write enough there; as sketches they are pretty good (but still somewhat of a challenge to fill in; less of a burden these days for beginners since now FGA Explained exists). –  BCnrd Sep 30 '10 at 3:02

I can't give the answer right off, but a reference should be Koll{\'a}r's book "Rational Curves on Algebraic Varieties." Here's where he proves that it's a representable functor, and I believe that the lemma on the next page is what says "open subfunctor", though I might be off.

share|improve this answer
    
The relevant algebra fact is Proposition I.7.4.1 –  mdeland Sep 29 '10 at 22:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.