Exercise 13.3.8

Proof.

(a)

Let $\phi =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4)$. We know by Ex. 13.3.4(a) that the orbit ${\mathcal{}O}_{\phi }=S_n\cdot \phi$ is

$${\mathcal{}O}_{\phi }=\{{\phi }_0,{\phi }_1,\dots ,{\phi }_5\},$$

where

$$\begin{array}{llll}\hfill {\phi }_0=()\cdot \phi & =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4),& \hfill & \\ \hfill {\phi }_1=(34)\cdot \phi & =\sqrt{\mathrm{\Delta }}(-x_1-x_2+x_3+x_4),& \hfill & \\ \hfill {\phi }_2=(23)\cdot \phi & =\sqrt{\mathrm{\Delta }}(-x_1+x_2-x_3+x_4),& \hfill & \\ \hfill {\phi }_3=(14)\cdot \phi & =\sqrt{\mathrm{\Delta }}(x_1-x_2+x_3-x_4),& \hfill & \\ \hfill {\phi }_4=(13)\cdot \phi & =\sqrt{\mathrm{\Delta }}(x_1-x_2-x_3+x_4),& \hfill & \\ \hfill {\phi }_5=(24)\cdot \phi & =\sqrt{\mathrm{\Delta }}(-x_1+x_2+x_3-x_4).& \hfill & \\ \hfill & & \hfill \end{array}$$

We verify that the orbit $A_n\cdot \phi$ is ${\mathcal{}O}_{\phi }$. Since $A_n$ fixes $\sqrt{\mathrm{\Delta }}$, we obtain

$$\begin{array}{cc}\hfill \sigma \hfill & \hfill \sigma \cdot \phi \hfill \\ & \\ \hfill ()\hfill & \hfill \sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4)\hfill \\ \hfill (13)(24)\hfill & \hfill \sqrt{\mathrm{\Delta }}(-x_1-x_2+x_3+x_4)\hfill \\ \hfill (124)\hfill & \hfill \sqrt{\mathrm{\Delta }}(-x_1+x_2-x_3+x_4)\hfill \\ \hfill (234)\hfill & \hfill \sqrt{\mathrm{\Delta }}(x_1-x_2+x_3-x_4)\hfill \\ \hfill (243)\hfill & \hfill \sqrt{\mathrm{\Delta }}(x_1-x_2-x_3+x_4)\hfill \\ \hfill (134)\hfill & \hfill \sqrt{\mathrm{\Delta }}(-x_1+x_2+x_3-x_4)\hfill \\ & \\ \hfill \hfill \end{array}$$

This proves $A_n\cdot \phi \supset S_n\cdot \phi$, and since $A_n\subset S_n$, $A_n\cdot \phi \subset S_n\cdot \phi$, so that

$$A_n\cdot \phi =S_n\cdot \phi .$$

Here $G_f=A_n$ or $G_f=S_n$. Therefore $G_f$ acts transitively over the roots ${\phi }_i$ of the universal resolvent $\Theta (y)=\underset{i=0}{\overset{5}{\prod <!--\; FUNCTION\; APPLICATION\; -->}}(y-{\phi }_i)$.

Write $L=F({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$ be the splitting field of $f$, and $\beta ={\phi }_i({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)\in L,{\beta }_i=\phi ({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)\in L$.

Since ${\Theta }_f(y)=\underset{i=0}{\overset{5}{\prod <!--\; FUNCTION\; APPLICATION\; -->}}(y-{\phi }_i({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4))\in F[y]$, $\mathrm{Gal}(L∕F)$ acts on the roots of ${\Theta }_f(y)$. We show that this action is transitive.

We know that $A_n\cdot \phi =\{{\phi }_0,\dots ,{\phi }_5\}$, so that for each subscript $i\in \{0,\dots ,5\}$, there exists $\tau \in A_n$ such that $\tau \cdot \phi ={\phi }_i$.

Since $G_f\supset A_n$, $\tau \in G_f$, thus there is a corresponding $\sigma \in \mathrm{Gal}(L∕F)$ such that $\sigma ({\alpha }_j)={\alpha }_{\tau (j)},j=1,2,3,4$. Then

$$\begin{array}{llll}\hfill \sigma (\beta )& =\sigma (\phi ({\alpha }_1,\dots ,{\alpha }_4))& \hfill & \\ \hfill & =\phi (\tau ({\alpha }_1),\dots ,\tau ({\alpha }_5))& \hfill & \\ \hfill & =(\tau \cdot \phi )({\alpha }_1,\dots ,{\alpha }_4)& \hfill & \\ \hfill & ={\phi }_i({\alpha }_1,\dots ,{\alpha }_4)& \hfill & \\ \hfill & ={\beta }_i.& \hfill & \end{array}$$

Therefore the orbit of $\beta$ under the action of $\mathrm{Gal}(L∕F)$ is $\{\beta ={\beta }_0,\dots ,{\beta }_5\}$.

Then we prove that ${\Theta }_f(y)$ is irreducible with the same argument as in the proof of Theorem 7.1.1 or Proposition 6.3.7.

If $h$ is the minimal polynomial of $\beta$ over $F$, then $h$ divides ${\Theta }_f(y)$. $h\in F[y]$, and $\beta$ is a root of $h$, therefore ${\beta }_i=\sigma (\beta ),\sigma \in \mathrm{Gal}(L∕F)$ is a root of $h$. The six ${\beta }_i$ are distinct, as ${\Theta }_f(y)$ is separable, so that $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h)\ge 6$, where $h\mid {\Theta }_f$, thus ${\Theta }_f=h$ is irreducible.

(b)

Write, as in part (a), $$\begin{array}{llll}\hfill \phi & =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4),& \hfill & \\ \hfill {\phi }_2& =(124)\cdot \phi =\sqrt{\mathrm{\Delta }}(-x_1+x_2-x_3+x_4),& \hfill & \\ \hfill {\phi }_4& =(243)\cdot \phi =\sqrt{\mathrm{\Delta }}(x_1-x_2-x_3+x_4).& \hfill & \end{array}$$

Then the orbit of $\phi$ is $S_n\cdot \phi =\{\phi ,-\phi ,{\phi }_2,-{\phi }_2,{\phi }_4,-{\phi }_4\}$, and the universal resolvent is

$$\Theta (y)=(y^2-{\phi }^2)(y^2-{\phi }_2^2)(y^2-{\phi }_4^2).$$

Here $G_f=⟨(1324),(12)⟩$ is the dihedral group $D_8$. We compute the orbits of $\phi$ and of ${\phi }_2$ under the action of $G_f$.

$$\begin{array}{llll}\hfill (1324)\cdot \phi & =(-\sqrt{\mathrm{\Delta })}(x_3+x_4-x_2-x_1)& \hfill & \\ \hfill & =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4)=\phi & \hfill & \\ \hfill (12)\cdot \phi & =(-\sqrt{\mathrm{\Delta }})(x_2+x_1-x_3-x_4)=-\phi ,& \hfill & \end{array}$$

so that $G_f\cdot \phi =\{\phi ,-\phi \}$.

Write $\beta =\phi ({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)=\sqrt{\mathrm{\Delta }(f)}({\alpha }_1+{\alpha }_2-{\alpha }_3-{\alpha }_4)$, and similarly ${\beta }_2={\phi }_2({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4),{\beta }_4={\phi }_4({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$.Then

$${\Theta }_f(y)=(y^2-{\beta }^2)(y^2-{\beta }_2^2)(y^2-{\beta }_4^2).$$

If $\sigma \in \mathrm{Gal}(L∕F)$, and if $\tau \in G_f$ is the corresponding permutation, then, as in part (a),

$$\sigma (\beta )=(\tau \cdot \phi )({\alpha }_1,\dots ,{\alpha }_4)=\pm \phi ({\alpha }_1,\dots ,{\alpha }_4)=\pm \beta .$$

We know that $\beta \ne 0$, as ${\Theta }_f$ is separable. Thus $\sigma (\beta )\ne \beta$, so that $\beta \notin F$, but $\sigma ({\beta }^2)={(\pm \beta )}^2={\beta }^2$, thus ${\beta }^2\in F$. This proves that the factor $g(y)=y^2-{\beta }^2\in F[y]$, and $y^2-{\beta }^2$ is irreducible over $F$.

Moreover,

$$\begin{array}{llll}\hfill (1324)\cdot {\phi }_2& =(-\sqrt{\mathrm{\Delta }})(-x_3+x_4-x_2+x_1)& \hfill & \\ \hfill & =\sqrt{\mathrm{\Delta }}(-x_1+x_2+x_3-x_4)& \hfill & \\ \hfill & =-{\phi }_4& \hfill & \\ \hfill {(1324)}^2\cdot {\phi }_2& =-(1324)\cdot {\phi }_4& \hfill & \\ \hfill & =-(-\sqrt{\mathrm{\Delta }})(x_3-x_4-x_2+x_1)& \hfill & \\ \hfill & =\sqrt{\mathrm{\Delta }}(x_1-x_2+x_3-x_4)& \hfill & \\ \hfill & =-{\phi }_2& \hfill & \\ \hfill {(1324)}^3\cdot {\phi }_2& =-(1324)\cdot {\phi }_2={\phi }_4& \hfill & \\ \hfill (12)\cdot {\phi }_2& =(-\sqrt{\mathrm{\Delta }})(-x_2+x_1-x_3+x_4)& \hfill & \\ \hfill & =\sqrt{\mathrm{\Delta }}(-x_1+x_2+x_3-x_4),& \hfill & \end{array}$$

so that $G_f\cdot {\phi }_2=\{{\phi }_2,-{\phi }_2,{\phi }_4,-{\phi }_4\}$. Therefore the orbit of ${\beta }_2$ under the action of $\mathrm{Gal}(L∕F)$ is $\{\pm {\beta }_2,\pm {\beta }_4\}$.

Write $h(y)=(y^2-{\beta }_2^2)(y^2-{\beta }_4^2)$. Since $h(y)={\Theta }_f(y)∕g(y)$, $h(y)\in F[y]$. The action of $\mathrm{Gal}(L∕F)$ on the roots of the separable polynomial $h$ is transitive, thus, as in part (a), $h$ is irreducible over $F$.

${\Theta }_f(y)=g(y)h(y)$, where $g(y),h(y)\in F[y]$ are irreducible of degree 2 and 4 respectively.

(c)

By computing the orbits of $\phi ,{\phi }_2,{\phi }_4$, we obtain $$\begin{array}{llll}\hfill (12)(34)\cdot \phi & =\phi ,& \hfill & \\ \hfill (13)(24)\cdot \phi & =-\phi ,& \hfill & \\ \hfill (12)(34)\cdot {\phi }_2& =-{\phi }_2,& \hfill & \\ \hfill (13)(24)\cdot {\phi }_2& =-{\phi }_2,& \hfill & \\ \hfill (12)(34)\cdot {\phi }_4& =-{\phi }_4,& \hfill & \\ \hfill (13)(24)\cdot {\phi }_4& =-{\phi }_4.& \hfill & \\ \hfill & & \hfill \end{array}$$

Therefore

$$G_f\cdot \phi =\{\phi ,-\phi \},G_f\cdot {\phi }_2=\{{\phi }_2,-{\phi }_2\},G_f\cdot {\phi }_4=\{{\phi }_4,-{\phi }_4\},$$

and the orbits of $\beta ,{\beta }_2,{\beta }_4$ under the action of $G=\mathrm{Gal}(L∕F)$ are

$$G\cdot \beta =\{\beta ,-\beta \},G\cdot {\beta }_2=\{{\beta }_2,-{\beta }_2\},G\cdot {\beta }_4=\{{\beta }_4,-{\beta }_4\}.$$

If $\sigma \in \mathrm{Gal}(L∕F)$ corresponds to $(12)(34)$, then $\sigma (\beta )=-\beta \ne \beta$, and $\sigma ({\beta }^2)={\beta }^2$, so that $\beta \notin F,{\beta }^2\in F$, and the factor $y^2-{\beta }^2\in F[y]$ is irreducible over $F$. Similarly, the two other factors are irreducible in $F[y]$.

${\Theta }_f(y)=(y^2-{\beta }^2)(y^2-{\beta }_2^2)(y^2-{\beta }_4^2)$ is a decomposition in three irreducible factors of $F[y]$.

(d)

Suppose now that $G_f=⟨(1324)⟩$. We know that $$\begin{array}{lll}\hfill (1324)\cdot \phi =\phi ,& & \hfill \end{array}$$

thus the corresponding $\sigma \in \mathrm{Gal}(L∕F)$ fixes $\beta =\phi ({\alpha }_1,\dots ,{\alpha }_4)$, so that $\beta \in F$.

$g_1(y)=y-\beta ,g_2(y)=y+\beta$ are two factors of degree 1 of ${\Theta }_f(y)$.

Moreover, we know from part (b) that

$$\begin{array}{lll}\hfill (1324)\cdot {\phi }_2=-{\phi }_4,{(1324)}^2\cdot {\phi }_2=-{\phi }_2,{(1324)}^3\cdot {\phi }_2={\phi }_4& & \hfill \end{array}$$

As in part $b$, $\mathrm{Gal}(L∕F)$ acts transitively on the distinct roots of $(y^2-{\beta }_2^2)(y^2-{\beta }_4^2)\in F[y]$, thus this polynomial is irreducible over $F$:

${\Theta }_f(y)=g_1(y)g_2(y)g_3(y)$, where $g_1(y),g_2(y),g_3(y)\in F[y]$ are irreducible of degree 1,1 and 4 respectively.

(e)

By Theorem 13.1.1, $G_f$ is conjugate to a permutation group examined in parts (a),(b), (c) or (d), and the four conclusions are mutually exclusive, so the factorisation of ${\Theta }_f(y)$ over $F$ enables to determine this case. In part (a), $\mathrm{\Delta }(f)$ enables in characteristic $\ne 2$ to distinguish the cases $G_f=A_4$ and $G_f=A_4$. Therefore the factorisation of ${\Theta }_f(y)$ and $\mathrm{\Delta }(f)$ give an algorithm to determine $G_f$ up to conjugacy.