# Questions about a Zariski open set II

In the paper Solutions to the XXX type Bethe ansatz equations and flag varieties, page 6, line 14, it is said that generic polynomials with respect to $\mathbf{z}$, $\mathbf{\Lambda}$ form a Zariski open subset of the population. Here generic polynomials are defined in page 5, line 6 and population is defined in page 6, line 11.

I think that if we want to show that the set of generic polynomial is a Zariski open set, then we have to show that its complement is defined by a set of equations. But I don't know how to show that its complement is defined by a set of equations.

Thank you very much.

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Let $u$ be a polynomial in a complex vector space $V$ which consists of polynomials. The coordinates of $u$ is represented by its coefficients. Suppose that $u=\prod_{i=1}^{n} (x-t_i)$. Then the coordinates of $u$ is $(1, -\sum_{i} t_i, \sum_{i<j}t_it_j, \ldots, (-1)^{n}t_1\cdots t_n)=(1, -\sigma_1, \ldots, (-1)^n\sigma_n)$ which consists of elementary symmetric functions.
We will show that the set of polynomials which have multiple roots is defined by a set of equations. It is sufficient to show that if $u$ has multiple roots, then the coordinates of $u$ satisfy an equation. $\prod_{i<j}(t_i-t_j)^2$ is a symmetric function and hence we have $\prod_{i<j}(t_i-t_j)^2=f(\sigma_1, \ldots, \sigma_n)$ for some polynomial $f$. $u$ does not have multiple roots if and only if $\prod_{i<j}(t_i-t_j)^2\neq 0$. Therefore, if $u$ has multiple roots, then $f(\sigma_1, \ldots, \sigma_n)=0$.
We will show that the set of polynomials which have common roots between any two polynomials in the set is defined by a set of equations. It is sufficient to show that if $u$ and $v$ do not have common roots, then the coordinates of $u$ and $v$ satisfy an equations. Suppose that $u=\prod_{i=1}^n(x-t_i), v=\prod_{j=1}^{m}(x-s_j)$. Then if $u, v$ do not have common roots, then $\prod_{i,j}(t_i-s_j)\neq 0$. Since $\prod_{i,j}(t_i-s_j)$ is a symmetric function with respect to $t_i$ (or $s_j$), $\prod_{i,j}(t_i-s_j)=g(\sigma_1, \ldots, \sigma_n)$, where the coefficients of $g$ are functions of $s_j$ and $g$ is symmetric with respect to $s_j$. Therefore $$\prod_{i,j}(t_i-s_j)=g(\sigma_1, \ldots, \sigma_n)=h(\sigma_1, \ldots, \sigma_n, \sigma'_1, \ldots, \sigma'_m).$$ Here $\sigma_i, \sigma'_i$ are elementary symmetric functions of $t_i$ and $s_j$ respectively.