What do all the modules of dimension 1 over $S_n$ look like? I need to find (up to isomorphism) all the complex modules over $S_n$ that are 1-dimensional. In the past I've done this for a fixed $n$, but never in general..
 A: Let me provide a general framework for what Tobias is discussing. I assume that we're talking about $\mathbb{C}$ representations, for convenience.
Recall that if $G$ is a group, it's commutator subgroup, denoted $G'$, is the subgroup of $G$ generated by all commutators (i.e. elements of the form $ghg^{-1}h^{-1}$. The abelinization of $G$ is the quotient $G/G'$, which we shall denote $G^\text{ab}$. The abelinization satisfies the following universal property:

For any homomorphism $G\to A$, where $A$ is abelian, there exists a unique homomorphisms $G^\text{ab}\to A$ such that the following triangle commutes
  $$\begin{array}{ccc}G & \to & A\\\downarrow & \nearrow &\\ G^\text{ab} &  & \end{array}$$

So, note that this implies that the quotient map $\pi:G\to G^\text{ab}$ induces a bijection $\text{Hom}(G,A)\to \text{Hom}(G^\text{ab},A)$ by $f\mapsto f\circ\pi$. 
Let us denote the set of one-dimensional representations of a group $G$ by $\text{Char}(G)$. It is clear that $\text{Char}(G)$ is really just the set $\text{Hom}(G,S^1)$, where $S^1$ is the circle group (a priori it
s $\text{Hom}(G,\mathbb{C}^\times)$, but the image of any map $G\to \mathbb{C}^\times$ needs to land in the torsion subgroup of $\mathbb{C}^\times$ which is contained in $S^1$). From the above then, we see that we obtain the following theorem

Theorem: The map $\text{Char}(G)\to\text{Char}(G^{\text{ab}})$ given by $f\mapsto f\circ\pi$ is a bijection.

So, let us consider your actual question. We seek to find $\text{Char}(S_n)$. Well, from the above it suffices to find $\text{Char}((S_n)^\text{ab})$. Now, it is somewhat of an interesting problem to show that $(S_n)'=A_n$. Thus, we see that $(S_n)^\text{ab}\cong\mathbb{Z}/2\mathbb{Z}$. From, this we deduce that $\text{Char}(S_n)$ is in one-to-one correspondence with $\text{Char}(\mathbb{Z}/2\mathbb{Z})$. But, it is trivial to verify that $\#\text{Char}(\mathbb{Z}/2\mathbb{Z})=2$.
Remark: In fact, note that $\text{Char}(A)$ for an abelian group is an abelian group since $\text{Hom}(A,S^1)$ is--with this observation, the following (true) statement makes sense: $\text{Char}(A)\cong A$ for any abelian group $A$. This is proven since trivially $\text{Hom}(\mathbb{Z}/n\mathbb{Z},S^1)\cong \mathbb{Z}/n\mathbb{Z}$ (since $S^1$ contains only one cyclic subgroup of order $n$--they are the solutions to $x^n=1$ (i.e. the $n^{\text{th}}$ roots of unity)), and then proceeding to the general case by applying the Fundamental Theorem for Finitely Generated Abelian Groups and the fact that Hom splits accross finite sums in the first entry. 
From this we may deduce that $\text{Char}(S_n)=2$, and we know precisely what they are. Namely, the two elements of $\text{Car}(\mathbb{Z}/2\mathbb{Z})$ are just $1\mapsto \pm 1$. Thus, following the map $\pi:S_n\to S_n/A_n=\{\pm 1\}$ we obtain our two maps--they are just $\pi$ itself and the zero map. But, $\pi$ is nothing but the sign map $\text{sgn}$. Thus, we obtain the final result, that $\text{Char}(S_n)=\{\chi^{\text{triv}},\chi^{\text{sgn}}\}$ where $\chi^{\text{triv}}(\sigma)=1$ for all $\sigma\in S_n$ and $\chi^{\text{sgn}}(\sigma)=\text{sgn}(\sigma)$ for all $\sigma\in S_n$.
Note: My usage of the notation $\text{Char}(G)$ is non-standard. The more conventional notation is $\widehat{G}$, and we call this object the dual group. I just didn't like the look of $\widehat{G^\text{ab}}$.
