Exercise 13.1.15

Proof. By (12.17), the roots of the quartic $f=x^4-c_1{x}^3+c_2{x}^2-c_{3}x+c_4$ are

where $y_1,y_2,y_3$ are the roots of the Ferrari resolvent

and the ${𝜀}_i=\pm 1$ are chosen so that the product of the radicals $t_i=+{𝜀}_i\sqrt{4y_i+c_1^2-4c_2}$ is

Let $L=F({\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4)$ be the splitting field of $F$.

Here ${𝜃}_f(y)$ has a root in $F$, say $y_1$. Thus

where $g(y)=y^2+\mathit{ay}+b\in F[y].$ Therefore the roots $y_2,y_3$ of $g$ are in $F(\sqrt{\delta })$, where $\delta =a^2-4b\in F$ is the discriminant of $g$. Moreover $t_1={\alpha }_1+{\alpha }_2-{\alpha }_3-{\alpha }_4=\sqrt{4y_1+c_1^2-4c_2}\in L$, and similarly $t_2,t_3\in L$, so $F(t_1,t_2,t_3)\subset L$, and by (12.17), $L\subset F(t_1,t_2,t_3)$, therefore

Since $\mathrm{\Delta }({𝜃}_f)=\mathrm{\Delta }(f)\ne 0$, there is at most one $t_i$ equal to $0$. So we can choose the numbering such that $t_1{t}_2\ne 0$ (perhaps $t_3=0$). Since $t_1{t}_2{t}_3=c_1^3-4c_1{c}_2+8c_3\in F$, $t_3\in F(t_1,t_2)$, so

Note that $t_i^2=4y_i+c_1^2-4c_2\in L$, so $y_i\in L,i=1,2,3$, so $\sqrt{\delta }=y_2-y_3\in L$, therefore $L(\sqrt{\delta })=L$. Consider the chain of inclusions

Since $4y_1+c_1^2-4c_2\in F,\delta \in F$ and $4y_2+c_1^2-4c_2\in F(\sqrt{\delta }$), the degree of each extension is 1 or 2, so

Moreover $L\supset F({\alpha }_1)$, and the minimal polynomial of ${\alpha }_1$ is f, so

Since $|G|=[L:F]$,

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