Let $K$ be an algebraic number field, $f(x) = 0$ an algebraic equation over $K$ of degree $n \ge 2$ with only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting field of this equation over $K$. Any $y \in L$ then has a representation $$y = P(x_1, \dotsc, x_n)\label{1}\tag{1}$$ with $P = P(X_1, \dotsc, X_n) \in K[X_1, \dotsc, X_n]$ a polynomial in the $n$ indeterminates $X_1, \dotsc, X_n$ over $K$. Such an element is also called a rational function of the roots — this being a priori a quotient $Q(x_1, \dotsc, x_n)/R(x_1,\dotsc, x_n)$, but such an expression can always be rewritten as a polynomial $P(x_1, \dotsc, x_n)$ —. A resolvent for the equation $f(x) = 0$ is then an expression $t = P(x_1, \dotsc, x_n)$ having, in classical parlance, the property that any rational function of the roots can be expressed as a rational function of $t$, or being, in modern terminology, a primitive element of the field extension $L:K$ (for this and what follows see [1]).
This notion goes back to Lagrange and Galois, and was taken by the latter as the point of departure of his theory of the solvability of equations (later vigorously exorcised by Dedekind and Artin, making the theory questionably more elegant and unquestionably much more impenetrable). The point of departure for obtaining, in return, such a resolvent, was the following observation. Fix once and for all an enumeration of the roots as $x_1, \dotsc, x_n$. Any permutation $\sigma \in \mathfrak{S}_n$ operates on the polynomials $P \in K[X_1, \dotsc, X_n]$ via $P \mapsto P^{\sigma}$ with $$P^{\sigma}(X_1, \dotsc, X_n) \mathrel{:=} P(X_{\sigma(1)}, \dotsc, X_{\sigma(1)}).$$ Thus to any $y \in L$ with a given representation \eqref{1} there is attached a bunch of at most $n!$ different values $y^{\sigma} \mathrel{:=} P^{\sigma}(x_1, \dotsc, x_n)$. There are two basic claims:
(C1) If the values $t^{\sigma}$ are pairwise different, so that there are exactly $n!$ different ones, then $t$ is a resolvent;
(C2) for every degree $d \ge 1$ (and even only $\mathbb{Z}$-coefficients), the generic polynomial $P$ gives $t = P(x_1, \dots, x_n)$ satisfying (C1), thus yielding a resolvent.
Here, "generic" means "not satisfying a certain nontrivial polynomial relation".
Now the condition (C1) is initially only sufficient, and the property of being a resolvent does not make use of the special representation \eqref{1}, so it is a priori possible that there is a representation $t = Q(x_1, \dots, x_n)$ not satisfying (C2) with $t$ still being a resolvent.
Question:
Can this happen?
[1] Edwards, H.M., Galois Theory (Graduate Texts in Mathematics 101). Springer 1984
