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] <cite authors="Edwards, H.M.">Edwards, H.M.,
_Galois  Theory_
(Graduate Texts in Mathematics 101). Springer 1984