Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a smooth projective and irreducible curve over a field $k$. Further, define $$ X_d = \{ \text{ Effective Cartier divisors of degree } d \text{ on } X \;\} $$ and $$ W_d = \{ \text{ Line bundles of degree } d \text{ on } X \;\}. $$ What is the best way to give the above sets a scheme structure and show that those schemes are irreducible?

I'm not sure, intuitively, if the irreducibility holds in general or if we need further assumptions on $X$and/or $k$.

PS: A way to give a scheme structure to $X_d$ is the following: Assume $X$ is a scheme over another scheme $S$ and consider the functor $$ Div^d_{X/S}: Sch_S \to Set, \quad T\mapsto \{ \text{ Relative eff Car divisors of deg } d \text{ on } X_T/T \; \} $$ and assume that it is representable (this is true for $X$ curve as above). Then one defines $X_d$ as the representing $S$-scheme.

We can define $W_d$ similarly with an appropriate functor $\;Sch_S \to Set$.

share|improve this question
    
I guess the answers should be in Kleiman's notes, arXiv:math/0504020. –  Adeel Dec 1 '13 at 15:20
    
I looked into it, but I didn't find it. I may have missed it, but a search of the term "irreducible" in the PDF doesn't seem to find any answer to my question –  Abramo Dec 1 '13 at 15:21
    
Have you checked out these notes: math.stanford.edu/~conrad/248BPage/handouts/pic.pdf? They might be helpful. –  Dori Bejleri Dec 1 '13 at 20:32
add comment

1 Answer 1

up vote 4 down vote accepted

Effective divisors of degree $d$ can be seen as the symmetric product $\mbox{Sym}^d(X):=X^d/S_d$ where $S_d$ is the symmetric group on $d$ letters. $X^d$ is irreducible since $X$ is, and so $X^d$ and $\mbox{Sym}^d(X)$ receive a natural structure of variety.

As far as $W_d$, identify $JX$, the Jacobian of $X$, with $\mbox{Pic}^0(X)$. Let $L_d$ be a line bundle of degree $d$ on $X$, and define the map $\mbox{Pic}^0(X)\to\mbox{Pic}^d(X)$ where $L\mapsto L\otimes L_d$ where $\mbox{Pic}^d(X)$ denotes the set of line bundles of degree $d$ on $X$. This is a (non-canonical) isomorphism, and since $JX$ is irreducible and has the structure of variety, we get the same for $\mbox{Pic}^d(X)$

share|improve this answer
    
Dear Robert, this is definitely the right idea. Just one thing: how do we know the quotient by the action of $S_d$ is a variety? –  Bruno Joyal Dec 2 '13 at 1:49
2  
For a smooth projective variety, the quotient by a finite group always exists. This can be seen by showing that every point has an affine neighborhood that is fixed by the group and so there we can take the quotient (it isn't difficult to show that the quotient exists in the affine case). Then all the affine patches of the quotient are glued together. –  Robert Auffarth Dec 2 '13 at 1:52
    
Cool, thank you! –  Bruno Joyal Dec 2 '13 at 1:58
    
Thanks a lot Robert! And how can one see that the Jacobian is irreducible under our hypothesis? –  Abramo Dec 2 '13 at 2:02
2  
No problem! The fact that the Jacobian is irreducible is a well-known fact since it is an abelian variety of the same dimension as the genus of $X$. But yes, it is also because $X_d\to W_d$ is surjective for $d\geq g$ (since for $d\ge g$ every line bundle of degree $d$ comes from an effective divisor. This can be seen with Riemann-Roch, for instance). For $d\leq g$, the map $\mbox{Sym}^d(X)\to W_d$ is a birational morphism with its image! –  Robert Auffarth Dec 2 '13 at 2:12
show 1 more comment

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.