Irreducibility of the space of divisors on a curve

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 $\;Sch_S \to Set$.

-
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

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 $\mbox{Pic}^d(X)$
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
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! – rfauffar Dec 2 '13 at 2:12