Primitive of a splitting field obtained from Galois resolvent I am reading Edwards' book on Galois theory and I am stuck on the proof of a theorem: 
Let $f$ be a nonconstant polynomial over $K$, let $L$ be a splitting field of $f$ over $K$, and let $\theta_{1}$, $\theta_{2}$, $\dotsc$, $\theta_{n}$ be the distinct roots of $f$ in $L$. There is a polynomial $V$ in $n$ variables over $L$ of the form
$$
V(x_{1},x_{2},\dotsc,x_{n})=c_{1}x_{1}+c_{2}x_{2}+\dotsb+c_{n}x_{n}
$$
where $c_{1}$, $c_{2}$, $\dotsc$, $c_{n}$ are members of $K$ and such that the $n!$ evaluations of the permutations of $(\theta_{1},\theta_{2},\dotsc,\theta_{n})$ under $V$ are all distinct. For any such evaluation $v$,
$$
L=K(v).
$$
In particular, every root of $f$ is a polynomial in $v$.
(This is not the same notation used by Edwards.) I understand the part of the proof in finding $V$. I am stuck trying to understand why every root of $f$ is a polynomial in $v$. Without loss of generality, take $v=V(\theta_{1},\theta_{2},\dotsc,\theta_{n})$, let $\sigma_{1}$, $\sigma_{2}$, $\dotsc$, $\sigma_{(n-1)!}$ be the permutations of order $n$ that leave $1$ fixed, and put
$$
F(x,y) = \prod_{i=1}^{(n-1)!} (x-V(y,\theta_{\sigma_{i}2},\theta_{\sigma_{i}3},\dotsc,\theta_{\sigma_{i}n})).
$$
The strategy seems to be to show that the g.c.d. of $f(y)$ and $F(v,y)$ in $K(v)$ is the linear polynomial $y-\theta_{1}$. It's clear to me that $\theta_{1}$ is a root of $F(v,y)$, but I don't understand the argument contained in the text afterwards, that is, I don't understand why $F(v,\theta_{2})\neq0$. 
 A: Your definition of $F(x,y)$ is wrong in a subtle way.  To define $F(x,y)$, we first consider the polynomial $$G(x,y,t_2,\dots,t_n)=\prod_{i=1}^{(n-1)!} (x-V(y,t_{\sigma_{i}2},t_{\sigma_{i}3},\dotsc,t_{\sigma_{i}n}))$$ where we use formal variables instead of the $\theta_i$.  Now, $G(x,y,t_2,\dots,t_n)$ is symmetric in $t_2,\dots,t_n$, so we can instead consider it as a polynomial in $x$, $y$, and the elementary symmetric functions of $t_2,\dots,t_n$.  If we add an additional variable $t_1$, then the elementary symmetric functions of $t_2,\dots,t_n$ can be expressed as polynomials in $t_1$ and the elementary symmetric functions in $t_1,\dots,t_n$ (by performing polynomial division of $\prod_{i=1}^n(x-t_i)$ by $x-t_1$ to get the coefficients of $\prod_{i=2}^n(x-t_i)$, for instance).  So we can write $$G(x,y,t_2,\dots,t_n)=H(x,y,t_1,s_1,\dots,s_n)$$ for some polynomial $H$ where $s_1,\dots,s_n$ are the elementary symmetric functions in $t_1,\dots,t_n$.
Now we define $$F(x,y)=H(x,y,y,s_1(\theta_1,\dots,\theta_n),\dots,s_n(\theta_1,\dots,\theta_n)).$$ In other words, we substitute $y$ for the $t_1$ input to $H$, and then evaluate each of the elementary symmetric functions by substituting the $\theta_i$ for $t_i$.  These elementary symmetric functions are then elements of $K$, so $F(x,y)$ is a polynomial in $x$ and $y$ with coefficients in $K$.
(The difference between this definition and your definition $F(x,y)=G(x,y,\theta_2,\dots,\theta_n)$ is that we are substituting $y$ for all instances of $t_1$ that arise in writing $G$ in terms of $t_1$ and the symmetric functions $s_1,\dots,s_n$.  Your definition would substitute $\theta_1$ instead of $y$ for those instances of $t_1$, which doesn't matter if we're evaluating $F(x,\theta_1)$ but does matter if we're evaluating $F(x,\theta_k)$ for $k\neq 1$.)
Let's now trace through what we get when we evaluate $F(x,\theta_k)$ for some $k$.  Writing $\sigma_i=s_i(\theta_1,\dots,\theta_n)$, we get $$F(x,\theta_k)=H(x,\theta_k,\theta_k,\sigma_1,\dots,\sigma_n).$$  Now let me observe that if we set $t_1=\theta_k$ and $t_2,\dots,t_n$ to be all the $\theta_j$ for $j\neq k$ in some order, the $\sigma_i$ are the elementary symmetric functions in $t_1,t_2,\dots,t_n$.  Thus for this assignment of $t_2,\dots,t_n$, we have $$G(x,\theta_k,t_2,\dots,t_n)=H(x,\theta_k,\theta_k,\sigma_1,\dots,\sigma_n)=F(x,\theta_k).$$  Recalling the definition of $G$, this means $F(x,\theta_k)$ is the product $$\prod_{i=1}^{(n-1)!} (x-V(\theta_k,\theta_{\sigma_{i}2},\theta_{\sigma_{i}3},\dotsc,\theta_{\sigma_{i}n}))$$ where now $\sigma_i$ ranges over all bijections from $\{2,\dots,n\}$ to $\{1,\dots,n\}\setminus\{k\}$ (in other words, we are taking all permutations of the $\theta_j$ for $j\neq k$).  This means that $F(x,\theta_k)=0$ iff $x$ is $V$ evaluated at some permutation of the $\theta_i$ with $\theta_k$ coming first.  If $v=V(\theta_1,\dots,\theta_n)$, then, it is now clear that $F(v,\theta_1)=0$ and $F(v,\theta_k)\neq 0$ for $k\neq 1$.
