Irreducibility of Schur polynomials

Question

A natural question covering both this and this question would be

Let $n>2$. Describe Young diagrams $\lambda$ with at most $n$ nonempty rows (or equivalently non-increasing sequences $\lambda=(\lambda_1\ge\lambda_2\ge\ldots\ge\lambda_n\ge0$) for which the Schur polynomial $$s_\lambda(x_1,\ldots,x_n):=\frac{\det(x_i^{\lambda_j+n-j})}{\det(x_i^{n-j})}$$ is irreducible in $\mathbb{C}[x_1,\ldots,x_n]$.

Partial results:

1. Of course, for $\lambda_n>0$, this polynomial is divisible by $s_{1^n}=e_n$ , and so isn't irreducible. (This is a generalisation of the fact that $e_k=s_{1^k}$ can only be irreducible for $k< n$ (which is shown to be true in an answer to one of the questions linked above).

2. For $\lambda=(m)$, we have $s_\lambda=h_m$, and so one of the questions linked above shows that $s_\lambda$ is irreducible.

3. For $\lambda=(n-1,n-2,\ldots,0)$ we have $s_\lambda=\prod_{i< j}(x_i+x_j)$ by Vandermonde, which is very reducible.

Of course, it is mere curiousity forcing me to ask it, but I think that this question has a potential to have a meaningful answer.

Answer 1

Let $g=\gcd(\lambda_1+n-1,\lambda_2+n-2,\dots,\lambda_n)$. And let $V _g(X)=\prod _{i < j}\frac{x_i^g-x_j^g}{x_i-x_j}$. David already mentioned that $s_{\lambda}(X)$ is divisible by $V_g(X)$. This is in fact all that can happen.

Theorem: For every partition $\lambda$ (with $\lambda_n=0$), the polynomial $\frac{s_{\lambda}(X)}{V_g(X)}$ is either constant or irreducible.

This follows from theorem 3.1 in the paper "Newton functions generating symmetric fields and irreducibility of Schur polynomials" by Dvornicich and Zannier.

A result which is somewhat more general is stated and proved in "On the irreducibility of irreducible characters of simple Lie algebras". They extend the irreducibility above to characters of irreducible representations of simple Lie algebras divided by a Weyl denominator type factor (analog of $V_g$ above). Schur functions are a special case corresponding to $GL_n$.