Exercise 12.3.4

Proof.

(a)

Write $\beta =t_1{x}_1+\cdots +t_n{x}_n\in F[t_1,\dots ,t_n,x_1,\dots ,x_n]$, where $t_1,\dots ,t_n,x_1,\dots ,x_n$ are variables, and $$S(y)=\underset{\sigma \in S_n}{\prod }\left(y-(t_1{x}_{\sigma (1)}+\cdots +t_n{x}_{\sigma (n)}\right)=\underset{\sigma \in S_n}{\prod }(y-\sigma \cdot \beta ).$$

Then, for all $\tau \in S_n$,

$$\begin{array}{llll}\hfill \tau \cdot S(y)& =\tau \cdot \underset{\sigma \in S_n}{\prod }(y-\sigma \cdot \beta )& \hfill & \\ \hfill & =\underset{\sigma \in S_n}{\prod }(y-\tau \cdot (\sigma \cdot \beta ))& \hfill & \\ \hfill & =\underset{\sigma \in S_n}{\prod }(y-(\tau \circ \sigma )\cdot \beta ))& \hfill & \\ \hfill & =\underset{{\sigma }^{\prime }\in S_n}{\prod }(y-{\sigma }^{\prime }\cdot \beta )({\sigma }^{\prime }=\tau \circ \sigma )& \hfill & \\ \hfill & =S(y).& \hfill & \end{array}$$

By Theorem 2.2.7, $S(y)$ is a polynomial in $y$ with coefficients in $F[{\sigma }_1,\dots ,{\sigma }_n,t_1,\dots ,t_n]$, so

$$S(y)\in F[{\sigma }_1,\dots ,{\sigma }_n,t_1,\dots ,t_n,y].$$

The evaluation ${\sigma }_i\mapsto c_i,c_i\in F$ gives

$$s(y)\in F[t_1,\dots ,t_n,y].$$

Since the coefficients $a_i$ of $s(y)={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^{n!}{a}_i{y}^i$ are in the ring $F[t_1,\dots ,t_n]$, by (2.30),

$$\mathrm{\Delta }(s)=\mathrm{\Delta }(-a_1,\dots ,{(-1)}^i{a}_i,\dots ,a_{n!})\in F[t_1,\dots ,t_n].$$

(b)

By part (a), $s(y)\in F[t_1,\dots ,t_n,y]$, and $F\subset L$, so $s(y)\in L[t_1,\dots ,t_n,y]$. Moreover, since ${\alpha }_i\in L$, each factor of $s$ satisfies $$y-(t_1{\alpha }_{\sigma (1)}+\cdots +t_n{\alpha }_{\sigma (n)})\in L[t_1,\dots ,t_n,y].$$

(c)

Since $f$ is separable, the $n$ roots ${\alpha }_1,\dots ,{\alpha }_n$ of $f$ are distinct. Therefore, for all $\sigma ,\tau \in S_n$ such that $\sigma \ne \tau$, there exists some $i,1\le i\le n$ such that $\sigma (i)\ne \tau (i)$, so ${\alpha }_{\sigma (i)}\ne {\alpha }_{\tau (i)}$. Hence $$\sigma \ne \tau \Rightarrow t_1{\alpha }_{\sigma (1)}+\cdots +t_n{\alpha }_{\sigma (n)}\ne t_1{\alpha }_{\tau (1)}+\cdots +t_n{\alpha }_{\tau (n)}.$$

So $s(y)$ has $n!$ distinct roots. By Proposition 2.4.3, $\mathrm{\Delta }(s)$ is a nonzero element of $F[t_1,\dots ,t_n]$.