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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.

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