Problem. Find all elementary symmetric polynomials that are irreducible over $\mathbb{C}$.
My attempt.
It's easy to see that if we have polynomial $f(x_1, \dots, x_n)$ and it can be reduced to $g(x_1, \dots, x_n) \cdot h(x_1, \dots, x_n)$ then without loss of generality $g = g(x_1, \dots, x_k)$ and $h = h(x_{k+1}, \dots, x_n)$.
$g$ is symmetric towards $x_1, \dots, x_k$ and $h$ towards $x_{k+1}, \dots, x_n$.
Number of summands in $f(x_1, \dots, x_n)$ if $\deg(f) = l$ is $n \choose l$. Number of summands in $g, \deg(g) = s$ and $f$ are: $k \choose s$ and $n -k \choose l-s$. But ${n \choose l} > {k \choose s}{n -k \choose l-s}$ since there are more possibilities to choose $l$ elements from $n$ than to separate $n$ elements into $2$ groups of $k$ and $n-k$ elements and choose from one group $s$ elements and from the other $l-s$.
Let $e_k$ be the symmetric polynomial of degree $k$ in the variables $x_1, \dots, x_n$. Write $e_k$ as a product $cf_1 \dots f_r$, with each $f_i$ irreducible.
Since $e_k$ has degree one with respect to each variable, each variable must appear in precisely one of the factors $f_1, \dots, f_r$, resulting in a corresponding partition $E_1$, $E_2$, ..., $E_r$ of the set of variables. By the uniqueness of the decomposition, permuting the variables must result, essentially, in the same decomposition (with an irreducible factor at worst being multiplied by a nonzero constant), hence in the same partition of the variables.
This can be achieved in only two ways. Either $r = 1$, or else $r = n$ and each $E_i$ has cardinality one.
If $r = 1$, then $e_k$ is irreducible.
In the other case, the degree of $e_k$ must be at least $n$, so we must have $k = n$.