Exercise 13.3.4

Proof. (a) Write $H(\phi )$ the symmetry group of $\phi =\sqrt{\mathrm{\Delta }}(x_1+x_2-x_3-x_4)$.

Since $(1324)$ is an odd permutation, $(1324)\cdot \sqrt{\mathrm{\Delta }}=-\sqrt{\mathrm{\Delta }}$, so

Therefore

To prove the converse, we show that the orbit ${\mathcal{}O}_{\phi }$ of $\phi$ under the action of $S_n$ contains at least 6 elements.

Since the six elements in the right column are distinct (if the characteristic is not 2), we see that $|{\mathcal{}O}_{\phi }|\ge 6$.

The stabilizer $G_{\phi }$ of $\phi$ in $S_n$ is $H(\phi )$, which contains $⟨(1324)⟩$, therefore $|G_{\phi }|\ge 4$, and $|{\mathcal{}O}_{\phi }|=\frac{|S_n|}{|G_{\phi }|}\le \frac{24}{4}=6$, so $|{\mathcal{}O}_{\phi }|=6$ and $|H(\phi )|=|G_{\phi }|=4$, thus

and the orbit of $\phi$ under $S_4$ is given by ${\mathcal{}O}_{\phi }=\{{\phi }_0=\phi ,{\phi }_1,\dots ,{\phi }_5\}$, where

Moreover,

where

Thus

where $y_1=x_1{x}_2+x_3{x}_4$.

If we apply the permutation $(23)$ on this equality, we obtain

where $y_2=x_2{x}_4+x_1{x}_3$.

If we apply the permutation $(13)$ on the same equality, we obtain

where $y_3=x_4{x}_1+x_3{x}_2$.

Finally,

where $y_1=x_1{x}_2+x_3{x}_4,y_2=x_1{x}_3+x_2{x}_4,y_3=x_1{x}_4+x_2{x}_3$ are the roots of the universal Ferrari resolvent $𝜃(y)$. (c) Let ${\Theta }_f(y)$ is obtained by specializing the resolvent $\Theta (y)$ of part (b) to $f=x^4+b{x}^2+d$. Then $x_i$ maps to the root ${\alpha }_i,i=1,\dots ,4$, ${\sigma }_1$ maps to $0$, ${\sigma }_2$ to $b$, $\mathrm{\Delta }$ to $\mathrm{\Delta }(f)$, and $y_i$ to ${\beta }_i$,where

are the roots of the Ferrari resolvent of $x^4+b{x}^2+d$.

We obtain

If we write for simplicity

then

We know from Exercise 13.1.10 that the Ferrari resolvent of $x^4+b{x}^2+d$ is

The substitution $y\mapsto \frac{u}{v}$ gives