Let $K$ be an algebraic number field, $f(x) = 0$ a separable algebraic equation over $K$ of degree $n \ge 2$, i.e. having only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting field of this equation over $K$. The basic construction on which Galois built his theory of equations was the following description of the splitting field $L$. Let $T \in K[X_1, \dots, X_n]$ be a Galois resolvent, i.e. a polynomial with the property that all the rational functions $\tau^{\sigma} := T(x_{\sigma(1)}, \dots, x_{\sigma(n)})$ in the roots of $f(x)$ are pairwise different elements of $L$ for all $n!$ permutations $\sigma \in \mathfrak{S}_n$ (such Galois reolvents exist and are in some sense not specified here any further "generic", see [1]). Then all the $\tau^{\sigma}$ are primitive elements in the sense that they generate $L$ over $K$, i.e. $L=K[\tau^{\sigma}]$; this is one of the first basic results of Galois.

In particular, put $\tau := \tau^{\textrm{id}}$. Then form the following polynomial of degree $n!$ over $K$: $$ F(T) := \prod_{\sigma \in \mathfrak{S}_n} (T-\tau^{\sigma}) \in K[T]. $$ Factor $F(T)$ into irreducibles $$ F(T) = \prod_{i=1}^k G_i(T), $$ and let $G(T):=G_j(T)$ be a factor with $G_i(\tau)=0$. Then $L \cong K[T]/(G(T))$(see [1]).

Question:

Are the $G_i(T)$ pairwise different, all of the same degree?

(In fact, all of the same degree should be evident.)

[1] Edwards, H.M., Galois Theory (Graduate Texts in Mathematics 101). Springer 1984