Why is the trace submodule of $FS_n$ the unique $1$-dimensional submodule? Suppose $V$ is an $n$-dimensional vector space over a field $F$, with basis $e_1,\dots,e_n$. Then $S_n$ acts on $V$ by the action $\sigma\cdot e_i=e_{\sigma(i)}$, and $V$ is a $FS_n$-module, where $FS_n$ is the group ring over $F$. 
The set of $F$-multiples of the vector $e_1+\cdots+e_n$ is $S_n$ invariant, hence a $1$-dimensional $FS_n$-submodule of $V$. If $n\geq 3$, why is this the unique $1$-dimensional submodule of $V$? This is mentioned in Example 10 on page 846 of Dummit and Foote.
I know that if $v\in V$ is fixed by all $\sigma\in S_n$, then $v$ is a multiple of $e_1+\cdots+e_n$. I take $M=\langle v\rangle$ to be an arbitrary $1$-dimensional $FS_n$-submodule. I know this means $M$ is $S_n$-stable, so for arbitrary $cv\in M$, $\sigma\cdot cv=c\sigma\cdot v=cdv$ for some $d$, since $\sigma\cdot v\in M$. 
If $d=1$, then it follows that every element of $M$ is fixed by the action of $S_n$, hence contained in the trace submodule, hence equal to it by dimension. I guess it would be enough to show the generator $v$ is fixed by the action of $S_n$, but I don't see how.
 A: We can argue as follows.
Let $n\geq 3$ and $V$ be an $FS_n$-module over a field $F$ with $\dim V = n$ and basis $\mathcal B = \{e_1,\ldots,e_n\},$ just as described in the question.
Suppose that $0\neq a\in V$ spans a $1$-dimensional $FS_n$-submodule of $V,$ i.e. $\dim FS_na = 1.$
For any $f\in V,$ let's denote by $f_j$ the coefficient of $e_j$ when expressing $f$ with respect to the basis $\mathcal B,$ i.e.
$$
f = f_1e_1 + \cdots f_ne_n.
$$
Observe that, for any $\lambda\in F,$ we have by linearity
$$
(\lambda f)_j = \lambda (f_j).
$$
Pick two different numbers $r,s\in\{1,\ldots,n\}$ and consider the transposition $(r,s) \in S_n.$ Please check the following facts.
$$
\begin{align}
(i) & \qquad((r,s)a)_j = a_j\ for\ r\neq j\neq s. &\\
(ii) & \qquad((r,s)a)_r = a_s. &\\
(iii)& \qquad((r,s)a)_s = a_r. &\\
\end{align}
$$
Moreover, since $\dim FS_na = 1,$ we must have
$$
(iv)\qquad(r,s)a = \lambda_{r,s} a\ for\ some\ \lambda_{r,s}\in F.
$$
Since $(r,s)$ is invertible and $a\neq 0,$ we must have
$$
(v)\qquad(r,s)a\neq 0,\ and\ thus\ \lambda_{r,s}\neq 0.
$$
The above considerations are valid for any pair $r,s$ of different indices.
Also, since $a\neq 0,$ there is an index $t$ such that
$$
(vi)\qquad a_t\neq 0.
$$
Now we pick an index $u\neq t$ and conclude, using (iii), (iv), (v), and (vi)
$$
0\neq a_t = ((t,u)a)_u = (\lambda_{t,u}a)_u = \lambda_{t,u}(a_u).
$$
This implies $a_u\neq 0.$ Since $u\neq t$ was arbitrary, we have shown that
$$
(vii)\qquad a_j\neq 0\ for\ every\ index\ j.
$$
Next, we pick indices $x,y$ with $x\neq y$ and $x\neq t\neq y.$ Using (iii) and (iv), we conclude
$$
a_x = ((x,y)a)_y = (\lambda_{x,y}a)_y = \lambda_{x,y} (a_y).
$$
But from (i), we get
$$
a_t = ((x,y)a)_t = (\lambda_{x,y}a)_t = \lambda_{x,y} (a_t).
$$
Since $a_t\neq 0,$ this implies $\lambda_{x,y} = 1,$ and thus
$$
a_x = a_y.
$$
Since the only conditions on $x$ and $y$ are $x\neq y$ and $x\neq t\neq y,$ we have shown the following.
$$
(viii)\qquad If\ a_t\neq 0,\ then\ for\ x\neq t\neq y: a_x=a_y.
$$
Using (vii), we can apply (viii) with $t = 1$ and $t=2,$ and get the following.
$$
(ix)\qquad For\ x\neq 1\neq y: a_x = a_y.
$$
$$
(x)\qquad For\ x\neq 2\neq y: a_x = a_y.
$$
Now we make use of $n\geq 3,$ which means we can (and will) consider the index $3.$ From (x), we conclude
$$
a_1 = a_3.
$$
From (ix), we conclude
$$
a_2 = a_3.
$$
From either (ix) or (x), we conclude
$$
\forall j\in \{4,\ldots n\}: a_j = a_3.
$$
All in all, we have shown that the coefficients $a_j$ are constant, independent of $j.$ That's exactly what we wanted.
