Exercise 13.1.9

Proof. If $\alpha$ is a root of $f$ in some splitting field $L$ of $F$, then ${\alpha }^4+a{\alpha }^3+b{\alpha }^2+\mathit{a\alpha }+1=0$. If we divide by ${\alpha }^4$, we obtain $1+a{\alpha }^{-1}+b{\alpha }^{-2}+a{\alpha }^{-3}+{\alpha }^{-4}$, so $f({\alpha }^{-1})=0$. Note that

As $0$ is not a root of $f$, the roots of $f$ are the roots of $z=x+\frac{1}{x}$, where $z$ is a root of $z^2+\mathit{az}+b-2$, so the roots of $f$ are the roots of the two polynomials

where $z_1,z_2$ are the roots in $L$ of

If we relabel the roots so that ${\alpha }_1,{\alpha }_2$ are the roots of $x^2-z_{1}x+1$, and ${\alpha }_3,{\alpha }_4$ the roots of $x^2-z_{2}x+1$, then ${\alpha }_1{\alpha }_2=1,{\alpha }_3{\alpha }_4=1$, therefore ${\beta }_1={\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4=2$ is a root of the Ferrari resolvent ${𝜃}_f(y)$. □