# What is known about zero-sets of Schur polynomials?

Consider a set S of partitions not containing the empty partition (I would be happy with, say, all the partitions of size less than k, except for the empty one).

Let $U_\lambda^{(r)}$ be the zero-set in $\mathbb{C}^r$ of the Schur polynomial $s_\lambda(x_1,\cdots,x_r)$.

What is known about $\cap_{\lambda \in S} U_\lambda^{(r)}$, beyond the fact that it is symmetric under the action of $S_r$?

(I am having trouble finding information about this: all the hits are about the different question of the Schur stability of univariate polynomials, a concept based on the location of the roots of those polynomials).

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For $k :=|\lambda| \ge r$, the statement that all $s_\lambda(x_1, \ldots, x_r)$ vanish is equivalent to all the elementary polynomials $e_j(x_1, \ldots, x_r) := \sum_{i_1 < \ldots < i_j} x_{i_1} \ldots x_{i_j}$, $j \le r$, vanish, since the latter form a basis of the algebra $\Lambda_r$. But this is true if and only if all the $x_1, \ldots, x_r$ are zero, since $e_j$ is the $y^{r-j}$ coefficient of a degree $r$ polynomial, i.e., $$\prod_{i=1}^r (y - x_i) = \sum_{j=0}^r (-1)^j e_j(x_1, \ldots, x_r) y^{r-j}.$$ If all the $e_j$'s are zero, except $e_0 \equiv 1$, then it must be $y^r$, hence all the $x_i$'s vanish.

For $k < r$, the zero set consists of roots of polynomials of the form $y^r + c_{k+1} y^{r-k-1} + c_{k+2} y^{r-k-2} + \ldots + c_r$.

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Assume there is a non-trivial common zero $\xi \neq (0,\dots,0)$.

Now, to quite wikipedia: "The degree d Schur polynomials in n variables are a linear basis for the space of homogeneous degree d symmetric polynomials in n variables."

So, let $P$ be any symmetric homogeneous polynomial in $n$ variables, of degree $d$. It is now clear that $P$ is a linear combination of the Schur polynomials, hence, $P$ must be zero at $\xi.$

Edit: This seems very strange, at least when the number of equations is greater than the number of variables. For small $r$, there might be a few zeros except 0, but when the number of partitions of $r$ is greater than $r$ itself, then one cannot apriori expect a solution except 0.

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Re: your 2rd paragraph. Only symmetric polynomials are linear combination of Schur polynomials. – Mariano Suárez-Alvarez Feb 8 '12 at 7:37
@Mariano Suárez-Alvarez : Ok, fixed. – Per Alexandersson Feb 8 '12 at 18:26