Just for curiosity, since I haven't found a justification around. Given a polynomial $P(x)=x^{n} − e_{1}x^{n−1} + e_{2}x^{n−2}- \dots + (−1)^{n}e_{n}$ of degree $n$ with discriminant $\Delta_x(P)=f(e_{1},e_{2}, \dots,e_{n})$, does there always exist a polynomial $Q(y)$ of degree $n-1$ with coefficients given by rational polynomials in $e_{1},e_{2}, \dots,e_{n}$ (in the roots $x_{1},x_{2},\dots,x_{n}$) such that $\Delta_y(Q)=\Delta_x(P)$?
Example: $P(x)=x^{3} − e_{1}x^{2} + e_{2}x-e_{3} \quad \Delta_x(P)=e_{1}^{2}e_{2}^{2}- 4e_{1}^{3}e_{3} – 4e_{2}^{3}+18e_{1}e_{2}e_{3} – 27e_{3}^{2}$
Is there $Q(y)=f_{2}(e_{1},e_{2},e_{3})y^2+f_{1}(e_{1},e_{2},e_{3})y+f_{0}(e_{1},e_{2},e_{3})$ such that $f_{1}^2-4f_{0}f_{2}=\Delta_x(P)$ with $f_{i}$ a rational polynomial over the $e$'s?
Update: The question is positive for $n=3$ or $n=4$, since solvability allows to construct a resolvent that has the same discriminant (or different by a constant factor). This has been asked in Why are the discriminants for a given polynomial & its resolvent polynomial congruent? and resolvent definitions appear in Resolvent cubic, Wikipedia; Kaplansky, Field and Rings; Algebra, Van der Waerden. Can we say that existence of $Q$ $\iff$ solvability?