Exercise 12.3.6

Proof.

(a)

Each term of $\tilde{B}$ has degree $n-1$ in $x$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\tilde{B})\le n-1$.

Let $\tau =(12)$. Then $\tau$ exchanges the two first terms of $\tilde{B}$,

$$\tau \cdot \left(\frac{\mathrm{\Delta }(x-x_2)(x-x_3)\cdots (x-x_n)}{{(x_1-x_2)}^2{(x_1-x_3)}^2\cdots {(x_1-x_n)}^2}\right)=\frac{\mathrm{\Delta }(x-x_1)(x-x_3)\cdots (x-x_n)}{{(x_2-x_1)}^2{(x_2-x_3)}^2\cdots {(x_2-x_n)}^2},$$

and fixes the other terms. Therefore $\tau \cdot \tilde{B}=\tilde{B}.$

Let $\sigma =(12\cdots n)$. Then

$$\begin{array}{llll}\hfill \sigma \cdot \left(\frac{\mathrm{\Delta }(x-x_2)(x-x_3)\cdots (x-x_n)}{{(x_1-x_2)}^2{(x_1-x_3)}^2\cdots {(x_1-x_n)}^2}\right)& =\frac{\mathrm{\Delta }(x-x_3)(x-x_4)\cdots (x-x_1)}{{(x_2-x_3)}^2{(x_2-x_4)}^2\cdots {(x_2-x_1)}^2}& \hfill & \\ \hfill & =\frac{\mathrm{\Delta }(x-x_1)(x-x_3)\cdots (x-x_n)}{{(x_2-x_1)}^2{(x_2-x_3)}^2\cdots {(x_2-x_n)}^2},& \hfill & \end{array}$$

so $\sigma$ maps the first term on the second, and similarly the second on the third,..., and the last on the first. Therefore $\sigma \cdot \tilde{B}=\tilde{B}$. Since $\sigma ,\tau$ are generators of the group $S_n$, every permutation of $S_n$ fixes $\tilde{B}$. So the coefficients of $\tilde{B}$ are symmetric polynomials in $x_1,\cdots ,x_n$. By Theorem 2.2.2,

$$\tilde{B}\in F[{\sigma }_1,\dots ,{\sigma }_n,x].$$

(b)

For each index $i$, $1\le i\le n$ $$\begin{array}{llll}\hfill \tilde{B}(x_i)& =\frac{\mathrm{\Delta }(x_i-x_1)\cdots (x_i-x_{i-1})(x_i-x_{i+1})\cdots (x_i-x_n)}{{(x_i-x_1)}^2\cdots {(x_i-x_{i-1})}^2{(x_i-x_{i+1})}^2\cdots {(x_i-x_n)}^2}& \hfill & \\ \hfill & =\frac{\mathrm{\Delta }}{(x_i-x_1)\cdots (x_i-x_{i-1})(x_i-x_{i+1})\cdots (x_i-x_n)}& \hfill & \\ \hfill & =\frac{\mathrm{\Delta }}{{\tilde{f}}^{\prime }(x_i)}& \hfill & \end{array}$$

So $\mathrm{\Delta }-\tilde{B}(x_i){\tilde{f}}^{\prime }(x_i)=0$: $\mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime }$ vanishes when $x=x_i$.

(c) By part (b), $x-x_i$ divides $\mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime }$ for each $i,1\le i\le n$. Since the polynomials $x-x_i,1\le i\le n$, are relatively prime, their product $\tilde{f}$ divides $\mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime }$ in $F[x_1,\dots ,x_n,x]$: $$\tilde{f}\mid \mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime }.$$

Set

$$Ã=\frac{\mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime }}{\tilde{f}}\in F[x_1,\dots ,x_n,x].$$

Then $Ã\tilde{f}+\tilde{B}{\tilde{f}}^{\prime }=\mathrm{\Delta }$.

Moreover, $\tilde{f}\in F[{\sigma }_1,\dots ,{\sigma }_n,x]$, therefore ${\tilde{f}}^{\prime }\in F[{\sigma }_1,\dots ,{\sigma }_n,x]$. Since every $\sigma \in S_n$ fixes $\mathrm{\Delta },\tilde{f},{\tilde{f}}^{\prime }$, $\sigma$ fixes $Ã$, so

$$Ã\in F[{\sigma }_1,\dots ,{\sigma }_n,x].$$

By part (a), $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\tilde{B})\le n-1$.

Since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\mathrm{\Delta })=0$, $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã\tilde{f})=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\mathrm{\Delta }-\tilde{B}{\tilde{f}}^{\prime })\le \mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\tilde{B}{\tilde{f}}^{\prime })=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\tilde{B})+\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->({\tilde{f}}^{\prime })\le (n-1)+(n-1)$. Therefore $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã)+n\le 2n-2$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã)\le n-2$.

$Ã,\tilde{B}$ have the desired properties:

There exist $Ã,\tilde{B}\in F[{\sigma }_1,\dots ,{\sigma }_n,x]$ such that

$$Ã\tilde{f}+\tilde{B}{\tilde{f}}^{\prime }=\mathrm{\Delta },\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(Ã)\le n-2,\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(\tilde{B})\le n-1.$$

□