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).
 A: 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
$y^r + c_{k+1} y^{r-k-1} + c_{k+2} y^{r-k-2} + \ldots + c_r$.
A: 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.
