Exercise 13.2.5

Proof.

$\text{∙}$

In the context of the proof of part (b) of Theorem 13.2.6, $A_5\subset G$. Let $${\tau }_1=e,{\tau }_2=(123),{\tau }_3=(234),{\tau }_4=(345),{\tau }_5=(145),{\tau }_6=(125),$$

and $h_i={\tau }_i\cdot h,i=1,\dots ,6$, where $h=u^2$ and

$$\begin{array}{llll}\hfill u=& x_1{x}_2+x_2{x}_3+x_3{x}_4+x_4{x}_5+x_5{x}_1& \hfill & \\ \hfill & -x_1{x}_3-x_3{x}_5-x_5{x}_2-x_2{x}_4-x_4{x}_1.& \hfill & \end{array}$$

By definition,

$${\beta }_i=h_i({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5).$$

Since $A_5\subset G$ and ${\tau }_i\in A_5$, there exists ${\sigma }_i\in \mathrm{Gal}(L∕F)$ such that ${\sigma }_i$ maps to ${\tau }_i$ for every $i\in \text{[[}1,6\text{]]}$, so

$${\sigma }_i({\alpha }_j)={\alpha }_{{\tau }_i(j)},i\in \text{[[}1,6\text{]]},j\in \text{[[}1,5\text{]]}.$$

Then, for all $i\in \text{[[}1,6\text{]]}$,

$$\begin{array}{llll}\hfill {\sigma }_i({\beta }_1)& ={\sigma }_i(h({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5))& \hfill & \\ \hfill & =h({\sigma }_i({\alpha }_1),{\sigma }_i({\alpha }_2),{\sigma }_i({\alpha }_3),{\sigma }_i({\alpha }_4),{\sigma }_i({\alpha }_5))& \hfill & \\ \hfill & =h({\alpha }_{{\tau }_i(1)},{\alpha }_{{\tau }_i(2)},{\alpha }_{{\tau }_i(3)},{\alpha }_{{\tau }_i(4)},{\alpha }_{{\tau }_i(5)})& \hfill & \\ \hfill & =({\tau }_i\cdot h)({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& \hfill & \\ \hfill & =h_i({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)& \hfill & \\ \hfill & ={\beta }_i& \hfill & \end{array}$$

So

$${\sigma }_i({\beta }_1)={\beta }_i,i=1,\dots ,6.$$

By assumption, some ${\beta }_j$ is in $F$, therefore ${\beta }_1={\sigma }_j^{-1}({\beta }_j)={\beta }_j\in F$, and ${\beta }_i={\sigma }_i({\beta }_1)={\beta }_1$ for all $i\in \text{[[}1,6\text{]]}$. We obtain the identity

$${(y-{\beta }_1)}^6={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y.$$

Multiplying this out, we obtain

$$\begin{array}{llll}\hfill & y^6-6{\beta }_1{y}^5+15{\beta }_1^2{y}^4-20{\beta }_1^3{y}^3+15{\beta }_1^4{y}^2-6{\beta }_1^{5}y+{\beta }_1^6=& \hfill & \\ \hfill & y^6+2b_2{y}^5+(b_2^2+2b_4)y^4+(2b_6+2b_2{b}_4)y^3+(b_4^2+2b_2{b}_6)y^2+(2b_4{b}_6-2^{10}\mathrm{\Delta }(f))y+b_6^2,& \hfill & \end{array}$$

so

$$\begin{array}{llll}\hfill -6{\beta }_1& =2b_2,& \hfill & \\ \hfill 15{\beta }_1^2& =b_2^2+2b_4,& \hfill & \\ \hfill -20{\beta }_1^3& =2b_2{b}_4+2b_6.& \hfill & \end{array}$$

Therefore, since $F$ has characteristic $\ne 2$,

$$\begin{array}{llll}\hfill b_2& =-3{\beta }_1,& \hfill & \\ \hfill b_4& =\frac{1}{2}(15{\beta }_1^2-9{\beta }_1^2)& \hfill & \\ \hfill & =3{\beta }_1^2,& \hfill & \\ \hfill b_6& =\frac{1}{2}(-20{\beta }_1^3+18{\beta }_1^3)& \hfill & \\ \hfill & =-{\beta }_1^3,& \hfill & \end{array}$$

so

$$b_2=-3{\beta }_1,b_4=3{\beta }_1^2,b_6=-{\beta }_1^3.$$

The precedent identity becomes

$$\begin{array}{llll}\hfill {(y-{\beta }_1)}^6& ={(y^3+b_2{y}^2+b_{4}y+b_6)}^2-2^{10}\mathrm{\Delta }(f)y& \hfill & \\ \hfill & ={(y^3-3{\beta }_1{y}^2+3{\beta }_1^{2}y-{\beta }_1^3)}^2-2^{10}\mathrm{\Delta }(f)y& \hfill & \\ \hfill & ={(y-{\beta }_1)}^6-2^{10}\mathrm{\Delta }(f)y.& \hfill & \end{array}$$

Hence $2^{10}\mathrm{\Delta }(f)=0$. Yet $F$ has characteristic $\ne 2$, and $\mathrm{\Delta }(f)\ne 0$, since $f$ is separable. This contradiction completes the proof of the theorem.

$\text{∙}$

We prove part (c) of Theorem 13.2.6.

Suppose that $G$ is conjugate to $⟨(12345)⟩$. Let $L=F({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4,{\alpha }_5)$ be the splitting field of $f$. We choose the numbering of the roots such that $\mathrm{Gal}(L∕F)=⟨\sigma ⟩$, where $\sigma$ corresponds to $(12345)$.

$[L:F]=|\mathrm{Gal}(L∕F)|=5$, so the Tower Theorem implies that $[F(\alpha ):F]$ divides 5.

Since $f$ is irreducible, $\alpha \notin F$, so $[F(\alpha ):F]\ne 1$, $[F(\alpha ):F]=5$ and $L=F(\alpha )$. Therefore ${\alpha }_i\in F(\alpha ),i=1,\dots ,5$, and so $f=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)(x-{\alpha }_4)(x-{\alpha }_5)$ splits completely over $F(\alpha )$.

Conversely suppose that $f$ splits completely over $F(\alpha )$. Then ${\alpha }_i\in F(\alpha ),i=1,\dots ,5$, so the splitting field of $f$ is $L=F(\alpha )$. Therefore $|\mathrm{Gal}(L∕F)|=[L:F]=5$ is prime, so $\mathrm{Gal}(L∕F)$ is cyclic. $G\simeq \mathrm{Gal}(L∕F)$ is cyclic of order 5, so $G=⟨\tau ⟩$, where $\tau \in S_5$ is a permutation of order 5. $\tau$ is a product of disjoint cycles whose order is the least common multiple of the length of the cycles, so $\tau$ is a $5$-cycle.

$G=⟨(a_1{a}_2{a}_3{a}_4{a}_5)⟩=\sigma ⟨(12345)⟩{\sigma }^{-1}$, where $\sigma$ is defined by $\sigma (i)=a_i,i=1,\dots ,5$. $G$ is conjugate to $⟨(12345)⟩$.