Exercise 12.3.8

Proof.

(a)

Let $$S(y)=\underset{\sigma \in S_n}{\prod }\left(y-(t_1{x}_{\sigma (1)}+\cdots +t_n{x}_{\sigma (n)})\right)=\underset{\sigma \in S_n}{\prod }(y-\sigma \cdot \beta ),$$

where $\beta =t_1{x}_1+\cdots +t_n{x}_n$.

As in (12.5), where here $\sigma \in S_n$ is the index of the sum, and $\psi =x_i,\phi =\beta ,{\phi }_{\sigma }=\sigma \cdot \beta ,{\psi }_{\sigma }=\sigma \cdot \psi =x_{\sigma (i)},𝜃=S$, define

$${\Phi }_i(y)=\underset{\sigma \in S_n}{\sum }\frac{S(y)x_{\sigma (i)}}{y-\sigma \cdot \beta }.$$

Since $\frac{S(y)}{y-\sigma \cdot \beta }=\underset{\tau \in S_n\setminus \{\sigma \}}{\prod <!--\; FUNCTION\; APPLICATION\; -->}(y-\tau \cdot \beta )$, ${\Phi }_i$ is a polynomial in $y$, with coefficients in $F[x_1,\dots ,x_n]$. Moreover, for all $\tau \in S_n$, since $\tau \cdot S(y)=S(y)$,

$$\begin{array}{llll}\hfill \tau \cdot {\Phi }_i(y)& =\underset{\sigma \in S_n}{\sum }\frac{S(y)x_{(\tau \circ \sigma )(i)}}{y-(\tau \circ \sigma )\cdot \beta }& \hfill & \\ \hfill & =\underset{{\sigma }^{\prime }\in S_n}{\sum }\frac{S(y)x_{{\sigma }^{\prime }(i)}}{y-{\sigma }^{\prime }\cdot \beta }({\sigma }^{\prime }=\tau \circ \sigma )& \hfill & \\ \hfill & ={\Phi }_i(y)& \hfill & \end{array}$$

Therefore,

$${\Phi }_i(y)\in F[{\sigma }_1,\dots ,{\sigma }_n,y].$$

If we evaluate the polynomial

$$\frac{S(y)}{y-\sigma \cdot \beta }=\underset{\tau \in S_n\setminus \{\sigma \}}{\prod }(y-\tau \cdot \beta )$$

at $\beta$, then we get ${\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \ne e}(\beta -\tau \cdot \beta )$ if $\sigma =e$ and $0$ otherwise. Therefore

$${\Phi }_i(\beta )=x_i\underset{\tau \ne e}{\prod }(\beta -\tau \cdot \beta ).$$

Moreover $S(y)={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\sigma \in S_n}(y-\sigma \cdot \beta )$, thus $S^{\prime }(\beta )={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\tau \ne e}(\beta -\tau \cdot \beta )$. We conclude, as in (12.8), that

$$x_i=\frac{{\Phi }_i(\beta )}{S^{\prime }(\beta )}.$$

(b)

The conclusion of Exercise 7 applied to $S=f$ shows that there are polynomials $A,B$ such that $$A(y)S(y)+B(y)S^{\prime }(y)=\mathrm{\Delta }(S).$$

Since the coefficients of $A$ and $B$ are polynomials in the coefficients of $S$, $A,B\in F[{\sigma }_1,\dots ,{\sigma }_n,y]$.

The definition of $S$ gives $S(\beta )=0$. Therefore

$$B(\beta )S^{\prime }(\beta )=\mathrm{\Delta }(S).$$

(c)

If we define ${\mathrm{\Psi }}_i(y)=B(y){\Phi }_i(y)$, then $$x_i=\frac{{\Phi }_i(\beta )}{S^{\prime }(\beta )}=\frac{B(\beta ){\Phi }_i(\beta )}{B(\beta )S^{\prime }(\beta )}=\frac{{\mathrm{\Psi }}_i(\beta )}{\mathrm{\Delta }(S)}.$$