"Unknot" algebraic set defined by two mutually dependent set of variables - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:59:49Z http://mathoverflow.net/feeds/question/66962 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66962/unknot-algebraic-set-defined-by-two-mutually-dependent-set-of-variables "Unknot" algebraic set defined by two mutually dependent set of variables Ewan Delanoy 2011-06-05T12:01:25Z 2011-06-05T19:47:31Z <p>Let $n$ be an integer $\geq 4$, and let $V \subseteq {\mathbb C}^{2n-1}$ be the set of all $(a_1,a_2, \ldots ,a_n,b_1,b_2, \ldots ,b_{n-1}) \in {\mathbb C}^{2n-1}$ such that the derivative of the polynomial $P=\prod_{k=1}^{n} (x-a_k)$ is exactly $n\prod_{k=1}^{n-1} (x-b_k)$. Then $V$ is a closed algebraic set in ${\mathbb C}^{2n-1}$ : it can be defined by the $n-1$ equalities </p> <p>$\sigma_k(b_1,b_2, \ldots ,b_{n-1})=\frac{n-k}{n}\sigma_k(a_1,a_2, \ldots ,a_{n-1},a_n) (1 \leq k \leq n-1)$,</p> <p>where $\sigma_k$ denotes the $k$-th symmetric polynomial.</p> <p>My question is, can $V$ be described as the image of a "rational" map, i.e. are there rational functions $f_1,f_2 \ldots f_{2n-1}$ in $r$ variables $y_1,y_2, \ldots ,y_r$ (where $r$ is another integer ) such that $V$ is exactly the image of the map $(f_1,f_2 \ldots f_{2n-1}): {\mathbb C}^{r} \to {\mathbb C}^{2n-1}$ ? Is $V$ a finite union of such images?</p>