Exercise 13.2.8

Proof.

(a)

Let $\tau \in S_5\setminus A_5$ be a transposition, and write $\sigma =(1243)\in S_5\setminus A_5$. We know that $\sigma \cdot u=-u$.

Since $\sigma \in S_5\setminus A_5$, $S_5$ is the disjoint union $S_5=A_5\cup A_5\sigma$, so $S_5\setminus A_5=A_5\sigma$. Since $\tau {\tau }_i\in S_5\setminus A_5$, then $\tau {\tau }_i\in A_5\sigma$, so

$$\tau {\tau }_i=\mathit{\psi \sigma },\psi \in A_5.$$

By Exercise 7, $\{{\tau }_1,{\tau }_2,\cdots ,{\tau }_6\}=\{e,(123),(234),(345),(145),(125)\}$ is a complete system of coset representatives of $\mathrm{AGL}(1,{𝔽}_5)\cap A_5$ in $A_5$. Therefore

$$\psi ={\tau }_j\phi ,j\in \text{[[}1,6\text{]]},\phi \in \mathrm{AGL}(1,{𝔽}_5)\cap A_5.$$

Since $\phi \in \mathrm{AGL}(1,{𝔽}_5)\cap A_5$, by Exercise 7 part (a), $\phi \cdot u=u$. Therefore

$$\begin{array}{llll}\hfill \tau \cdot u_i& =(\tau {\tau }_i)\cdot u& \hfill & \\ \hfill & =({\tau }_j\mathit{\phi \sigma })\cdot u& \hfill & \\ \hfill & =-({\tau }_j\phi )\cdot u=-{\tau }_{j}u=-u_j& \hfill & \end{array}$$

For each $i\in \text{[[}1,6\text{]]}$, there exists $j\in \text{[[}1,6\text{]]}$ such that $\tau \cdot u_i=-u_j$.

(b)

Let $$\Theta (y)=\underset{i=1}{\overset{6}{\prod }}(y-u_i)=y^6+B_1{y}^5+B_2{y}^4+B_3{y}^3+B_4{y}^2+B_{5}y+B_6.$$

Note that if $\tau \cdot u_i=\tau \cdot u_j$, $i,j\in \text{[[}1,6\text{]]}$, then ${\tau }^2\cdot u_i={\tau }^2\cdot u_j$, so $u_i=u_j$ and $i=j$. Therefore $\tau$ maps the set $\{u_1,\dots ,u_6\}$ on $\{-u_1,\dots ,-u_6\}$. Consequently

$$\begin{array}{llll}\hfill \tau \cdot \Theta (y)& =\underset{i=1}{\overset{6}{\prod }}(y-\tau \cdot u_i)& \hfill & \\ \hfill & =\underset{j=1}{\overset{6}{\prod }}(y+u_j)& \hfill & \\ \hfill & =y^6-B_1{y}^5+B_2{y}^4-B_3{y}^3+B_4{y}^2-B_{5}y+B_6& \hfill & \end{array}$$

Since

$$\tau \cdot \Theta (y)=y^6+\tau \cdot B_1{y}^5+\tau \cdot B_2{y}^4+\tau \cdot B_3{y}^3+\tau \cdot B_4{y}^2+\tau \cdot B_{5}y+\tau \cdot B_6,$$

we conclude

$$\tau \cdot B_i={(-1)}^i{B}_i,i=1,\dots ,6.$$

(c)

For $i=2,4,6$, $\tau \cdot B_i=B_i$ for every transposition $\tau$. Since every $\sigma \in S_5$ is a product of transpositions, for all $\sigma \in S_5$, $\sigma \cdot B_i=B_i$, where $B_i\in F[x_1,\dots ,x_5]$, therefore $B_i\in F[{\sigma }_1,\dots ,{\sigma }_5]$.

$B_2,B_4,B_6$ are polynomials in ${\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4,{\sigma }_5$.

(d)

For $i=1,3,5$, $\tau \cdot B_i=-B_i$, thus $B_i$ is invariant under $A_5$. Since the characteristic of $F$ is not 2, by Exercise 7.4.3, $B_i=C_i+D_i\sqrt{\mathrm{\Delta }}$, where $C_i,D_i$ are polynomials in the ${\sigma }_i$. Then $-C_i-D_i\sqrt{\mathrm{\Delta }}=-B_i=\tau \cdot B_i=C_i-D_i\sqrt{\mathrm{\Delta }}$, so $C_i=0$. $$B_i=D_i\sqrt{\mathrm{\Delta }},D_i\in F[{\sigma }_1,\dots ,{\sigma }_5]\text{ for }i=1,3,5.$$

(e)

Since $\sqrt{\mathrm{\Delta }}={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{1\le i<j\le 5}(x_i-x_j)$, $\sqrt{\mathrm{\Delta }}$ has degree $1+2+3+4=10$ as a polynomial in $x_1,x_2,x_3,x_4,x_5$. $B_1=u_1+u_2+u_3+u_4+u_5$, with $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(u_i)=2$, thus $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(D_1\sqrt{\mathrm{\Delta }})\le 2$. Therefore $D_1=0$.

$B_3=u_1{u}_2{u}_3+\cdots$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(B_3)=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(D_3\sqrt{\mathrm{\Delta }})\le 6$. Therefore $D_3=0$, and $B_3=0$.

$B_5=u_1{u}_2{u}_3{u}_4{u}_5+\cdots$, so $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(B_5)\le 10$. Therefore $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(D_5)\le 0$, so $D_5=c\in F$ is a constant.

$$\Theta (y)=y^6+B_2{y}^4+B_4{y}^2+B_6+c\sqrt{\mathrm{\Delta }}y,B_2,B_4,B_6\in F[{\sigma }_1,\dots ,{\sigma }_5],c\in F.$$

By Exercise 7, $c=-2^5$.