# Irreducibility of Schur polynomials

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.

• Basic observation: If $GCD(\lambda_1+n-1, \lambda_2+n-2, \dots, \lambda_{n-1}+1, \lambda_n) = g >1$, then $\prod_{i< j} (x_i^g-x_j^g)/(x_i-x_j)$ divides $s_{\lambda}$. – David E Speyer May 31 '12 at 18:07

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.
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 $GL_n$.
• Just for the sake of completeness: the answer to the original question is "$s_\lambda$ is irreducible if and only if $\lambda_n=0$ and $g=1$". – Vladimir Dotsenko Jun 1 '12 at 7:03
• I was looking at the paper by Dvornicich and Zannier as mentioned by Gjergji Zaimi above. I think you have stated the Theorem correctly, but there is one problem. If you read the proof carefully, you will find that the Theorem is proved, in one variable at a time, over other variables, using Galois theory. In simple word, irreducibility of polynomial $f$ in $\mathbb{C}[x_1,\dots,x_n]$ is different from irreducibility of $f$ in $x_n$ over $\mathbb{C}[x_1,\dots,x_{n-1}]$. Please check this out, and post if I am wrong. Thanks. – Neeraj Jun 21 '12 at 21:45