Exercise 12.2.2

Proof.

(a)

$(0,\dots ,0)\in W_{\sigma ,\tau }$. If $\lambda ,\mu \in F$ and $v=(t_1,\dots ,t_n)\in W_{\sigma ,\tau },w=(s_1,\dots ,s_n)\in W_{\sigma ,\tau }$, then ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})t_i=0$ and ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})s_i=0$, therefore $$0=\lambda \underset{i=1}{\overset{n}{\sum }}({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})t_i+\mu \underset{i=1}{\overset{n}{\sum }}({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})s_i=\underset{i=1}{\overset{n}{\sum }}({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})(\lambda t_i+\mu s_i),$$

so $\mathit{\lambda v}+\mathit{\mu w}\in W_{\sigma ,\tau }$. Thus $W_{\sigma ,\tau }$ is a subspace of $F^n$.

More shortly, $W_{\sigma ,\tau }$ is the kernel of the linear map

$$(t_1,\dots ,t_n)\mapsto \underset{i=1}{\overset{n}{\sum }}({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})t_i.$$

Since $\sigma \ne \tau$, there exists $k\in \text{[[}1,n\text{]]}$ such that $\sigma (k)\ne \tau (k)$. Moreover the ${\alpha }_i$ are distinct, so ${\alpha }_{\sigma (k)}\ne {\alpha }_{\tau (k)}$. Let $v=(t_1,\dots ,t_k,\dots ,t_n)=(0,\dots ,1,\dots ,0)$, where $t_k=1$ and $t_i=0$ if $i\ne k$. Then $v\in F^n$ satisfies ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n({\alpha }_{\sigma (i)}-{\alpha }_{\tau (i)})t_i={\alpha }_{\sigma (k)}-{\alpha }_{\tau (k)}\ne 0$, so $W_{\sigma ,\tau }\ne F^n$.

Therefore $W_{\sigma ,\tau }$ is a proper subspace of $F^n$, for all $\sigma ,\tau \in S_n$.

(b)

By Exercise 1, $$\underset{(\sigma ,\tau )\in S_n\times S_n,\sigma \ne \tau }{\bigcup }{W}_{\sigma ,\tau }\ne F^n.$$

Therefore there exists $(t_1,\dots ,t_n)\in F^n$ such that $(t_1,\dots ,t_n)\notin W_{\sigma ,\tau }$ for all $\sigma ,\tau \in S_n,\sigma \ne \tau$. This means that

$$\underset{i=1}{\overset{n}{\sum }}{\alpha }_{\sigma (i)}{t}_i\ne \underset{i=1}{\overset{n}{\sum }}{\alpha }_{\tau (i)}{t}_i,\text{ for all }\sigma ,\tau \in S_n,\sigma \ne \tau ,$$

so the $n!$ roots of

$$s(y)=\underset{\sigma \in S_n}{\prod }\left(y-(t_1{\alpha }_{\sigma (1)}+\cdots +t_n{\alpha }_{\sigma (n)})\right)$$

are distinct.