Example : Consider the (open, not compactified) moduli space of stable maps $ \mathcal M_g(1,d)$ of maps of smooth curves of genus $ g$ to $\mathbb P^1$. To each map we associate its branch divisor, which is an element of $ Sym^r(\mathbb P^1)$. Then we should have a morphism from $ \mathcal M_g(1,d)$ to $ Sym^r(\mathbb P^1)$. How do we prove this?

In general, if we have a family of objects $A$ and for each $ a \in A$ we can choose an element $ b \in B$ that "depends continuously" on $ a$. How do we prove that we have established a morphism $A \to B$? What is the general method to do this?

Note about the example: Extending the morphism to the Kontsevich compactification of the moduli space was the main objective of a paper by Fantechi-Pandharipande. But I couldn't filter out the proof that the map is a morphism, or the proof wasn't there.


It might help to identify $Sym^r(\mathbb P^1)$ with the Hilbert scheme of degree $r$ effective divisors on $\mathbb P^1$. The Hilbert scheme represents a functor, as does the space $\mathcal M_g(1,d)$, and so you can (try to) construct a map from one to the other by thinking in terms of the functors they represent.

As to why $Sym^r(C)$ is the Hilbert scheme of degree $r$ effective divisors when $C$ is a smooth projective curve, this is a (non-trivial, it has always seemed to me) exercise in making contact with reality from the somewhat more abstract world of flat families.

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    $\begingroup$ There is a nicely written rigorous explanation of the relationship between the $r$th symmetric power and the degree-$r$ Hilbert scheme of the curve in the discussion of Hilbert schemes and/or Jacobians in the book "Neron Models". $\endgroup$ – BCnrd Dec 8 '10 at 6:31

You have an answer to the 'example' version of your question already, but let me offer an answer to the "actual" question: if one is faced with two schemes $A$ and $B$, and for each $a\in A$ you have some way of constructing $f(a)\in B$, how might one check that $f$ is (or more precisely comes from) a morphism of schemes?

The answer, in many cases where $A$ is the solution to a moduli problem, is this. We're thinking of $A$ as parametrising objects $X$ (e.g. curves of genus $g$, elliptic curves with a point of order $n$ etc etc) and so for $a\in A$ you have some object $X_a$ corresponding to that point. You have a recipe that gives an element of $B$ (typically because $B$ is also the solution to a moduli problem) and you want to define the map $A\to B$ by following your nose.

But the insight is that the recipe you have, going from $A$ to $B$, might work in much more generality than you think. Let's take for example the map from the moduli space of elliptic curves plus points of order $n$, to the affine line, sending each elliptic curve to its $j$-invariant. This is "obviously" a continuous map $Y_1(n)\to{\mathbf{C}}$. But why is it a morphism of schemes?

[EDIT: I added the magic words "Weierstrass equation" to make this para correct] Well, if you go and read the definition of the $j$-invariant of an elliptic curve defined by a Weierstrass equation, then you see that if the coefficients of the Weierstrass equations are actually in a ring rather than a field, then the $j$-invariant of that curve is an element of that ring. Next one checks that $j$ is a well-defined invariant of the curve, that is, Weierstrass equations giving isomorphic curves have isomorphic $j$-invariants. But that solves your problem at a stroke! For say we have an $S$-valued point of $A$, for $S$ now any scheme. This corresponds to an elliptic curve over $S$. Now we can cover $S$ with affines such that on these affines the curve is defined by a Weierstrass equation. The $j$-invariant on these affines is a function on the affines, and uniqueness of $j$-invariants show that these functions glue (intersection of affines can be covered by affines---the usual trick) to get a well-defined function on the scheme $S$, that is, an $S$-valued point of the affine line. So for all $S$-valued points of $A$ we get an $S$-valued point of $B$ this way, just "following the definition" but applying it to the relative situation rather than the situation over fields. And the killer blow: now apply this to $S=A$, with the curve over $S$ equal to the universal curve over $A$. And there's your morphism.

[The above answer was initially too sloppy; thanks to Emerton and BCnrd for pointing this out below in the comments]

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    $\begingroup$ Dear Kevin, Regarding your $j$-invariant example, it depends on how you define elliptic curve over a ring. If you define it in terms of Weierstrass equations, then everything is okay. If you define it as a smooth proper family of genus one curves, then there is something to be proved to get a $j$-invariant which varies in families. (As far as I know, one has to first get a Weierstrass equation that is valid in the family.) Best wishes, Matt $\endgroup$ – Emerton Dec 8 '10 at 13:54
  • $\begingroup$ @Matt: I guess you're right! If I define it as a Weierstrass equation then everything is clear. But I guess it's not actually true that an elliptic curve over an affine scheme can globally be put into Weierstrass form. So I have to cover with smaller affines, where I can do it, and then glue the j-invariants. In other words the glueing procedure starts earlier than my answer indicates, doesn't it. $\endgroup$ – Kevin Buzzard Dec 8 '10 at 14:09
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    $\begingroup$ Dear Kevin: There are real problems if we define "elliptic curve" via Weierstrass cubics (rather than existence of such Zariski-locally on base being a thm; cf. Ch. 2 of Katz-Mazur). First, one still has to prove isoms respecting the identity section are given by the expected "change of variable" formulas (so $j$-invariant is really an invariant), which amounts to the same cohomological arguments use to construct Weierstrass eqns from the good defn. More seriously, can't work "locally" any finer than Zariski (e.g., etale, etc.), as needed to construct $Y_1(n)$! So best to use a good defn. $\endgroup$ – BCnrd Dec 8 '10 at 15:28
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    $\begingroup$ @Bcnrd: I absolutely agree. I was too quick to say it was "obvious" that $j$-invariant is well-defined for an ell curve over a ring. $\endgroup$ – Kevin Buzzard Dec 8 '10 at 19:53
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    $\begingroup$ The Weierstrass equations of elliptic curves over any base scheme $S$ (in particular the $j$-invariants) are considered in Deligne's paper "Courbes elliptiques: formulaire d'après J. Tate". Modular functions of one variable, IV (Proc. Internat. Summer School, Univ. Antwerp, Antwerp, 1972), pp. 53–73. Lecture Notes in Math., Vol. 476. $\endgroup$ – Qing Liu Dec 9 '10 at 10:39

Emerton has already answered, but let me summarize all the steps:

1) to a perfect torsion complex in the derived category associate a Cartier divisor;

2) to every sequence of morphisms $C\to X\to S$ satisfying suitable assumptions associate a perfect torsion complex, hence a Cartier divisor;

3) when $C\to X\to S$ is a family of stable maps to a smooth curve $Y$ (thus $X=Y\times S$), the divisor obtained on $X$ is effective and commutes with base change.

4) since $M_g(X,d)$ is the stack of stable maps, i.e. it represents a pseudofunctor (or, if you prefer, its coarse moduli space corepresents a functor), to define a morphism from it to a scheme $T$ means to define for every family over $S$ a mor $S\to T$ commuting with base change.

5) as Emerton wrote, $Hilb^rY=Sym^rY$. Hence an effective Cartier divisor on $X=S\times Y$ which is of degree $r$ on every fiber of $X\to S$ defines a morphism $S\to Sym^rY$.


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