Exercise 13.1.6

Proof.

(a)

If ${\alpha }_1+{\alpha }_2-{\alpha }_3-{\alpha }_4={\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4=0$, then $$\begin{array}{llll}\hfill s& :={\alpha }_1+{\alpha }_2={\alpha }_3+{\alpha }_4& \hfill & \\ \hfill p& :={\alpha }_1{\alpha }_2={\alpha }_3{\alpha }_4& \hfill & \end{array}$$

Thus $x^2-\mathit{sx}+p=(x-{\alpha }_1)(x-{\alpha }_2)=(x-{\alpha }_3)(x-{\alpha }_4)$, therefore

$$\{{\alpha }_1,{\alpha }_2\}=\{{\alpha }_3,{\alpha }_4\}.$$

Since ${\alpha }_3={\alpha }_1$ or ${\alpha }_3={\alpha }_2$, $f$ is not separable.

(b)

If $\beta$ is a root of the resolvent ${𝜃}_f$, we can relabel the roots of $f$ so that $\beta ={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4$ and $$4\beta +c_1^2-4c_2={({\alpha }_1+{\alpha }_2-{\alpha }_3-{\alpha }_4)}^2.$$

Since ${\beta }^2-4c_4={({\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4)}^2$, if $4\beta +c_1^2-4c_2$ and ${\beta }^2-4c_4$ both vanish, then ${\alpha }_1+{\alpha }_2-{\alpha }_2-{\alpha }_4=0$ and ${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4=0$. Then, by part (a), $f$ is not separable.

Therefore $4\beta +c_1^2-4c_2$ and ${\beta }^2-4c_4$ can’t both vanish when $f$ is separable.

(c)

Suppose that $4\beta +c_1^2-4c_2=0$ in part (c) of Theorem 13.1.1, where ${𝜃}_f(y)$ has a unique root $\beta$ in $F$. We know that $f$ is separable by Lemme 5.3.5.

Therefore $\mathrm{\Delta }(f)\ne 0$, and, by part (b), ${\beta }^2-4c_4\ne 0$, so

$$\mathrm{\Delta }(f)({\beta }^2-4c_4)\ne 0.$$

We know that $G=⟨(1324)⟩$ or $G=⟨(1324),(1,2)⟩$.

$\text{∙}$

Suppose that $G=⟨(1324)⟩$. Then $\mathrm{Gal}(L∕F)=⟨\sigma ⟩$, where $\sigma$ corresponds to $(1324)$. We choose $$\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)}=\sqrt{\mathrm{\Delta }(f)}({\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4).$$

Since $(1324)=(13)(32)(24)\notin A_4$, $\sigma (\sqrt{\mathrm{\Delta }(f)})=-\sqrt{\mathrm{\Delta }(f)}$, and

$$\sigma ({\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4)={\alpha }_3{\alpha }_4-{\alpha }_2{\alpha }_1=-({\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4).$$

Therefore $\sigma$ fixes $\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)}$, so $\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)}\in F^{\text{∗}}$, and

$$\mathrm{\Delta }(f)({\beta }^2-4c_4)\in {(F^{\text{∗}})}^2.$$

$\text{∙}$

Suppose that $G=⟨(1324),(1,2)⟩$. Then $\mathrm{Gal}(L∕F)=⟨\sigma ,\tau ⟩$, where $\tau$ corresponds to $(12)$. $\tau (\sqrt{\mathrm{\Delta }(f)})=-\sqrt{\mathrm{\Delta }(f)}$ and $\tau ({\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4)={\alpha }_2{\alpha }_1-{\alpha }_3{\alpha }_4={\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4$, so $\tau (\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)})=-\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)}$. Since the characteristic is not 2, and $\mathrm{\Delta }(f)({\beta }^2-4c_4)\ne 0$, $\sqrt{\mathrm{\Delta }(f)({\beta }^2-4c_4)}\notin F$, so $$\mathrm{\Delta }(f)({\beta }^2-4c_4)\notin {(F^{\text{∗}})}^2.$$

Therefore $G$ is conjugate to $⟨(1324),(12)⟩$ if and only if $\mathrm{\Delta }(f)({\beta }^2-4c_4)\notin {(F^{\text{∗}})}^2$.