Exercise 13.1.12

Proof.

(a)

Let $g=y^2+\mathit{Ay}+B\in F[y]$ and let $F\subset F(\sqrt{a}),a\in F$, be a quadratic extension.

If $\mathrm{\Delta }(g)=0$ then $a\mathrm{\Delta }(g)=0\in F^2$. Suppose now that $g$ is irreducible over $F$.

$\text{∙}$

Suppose that $g$ splits completely over $F(\sqrt{a})$, so $$g=(y-y_1)(y-y_2),y_1,y_2\in F(\sqrt{a}).$$

Then $\mathrm{\Delta }(g)={(y_1-y_2)}^2=A^2-4B\in F$. We choose $\sqrt{\mathrm{\Delta }(g)}=y_2-y_1\in F(\sqrt{a})$. Here $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(g)=2$, and $g$ is irreducible over $F$, therefore the roots of $g$

$$\begin{array}{llll}\hfill y_1& =\frac{1}{2}((y_1+y_2)+(y_1-y_2))=\frac{1}{2}\left(-A-\sqrt{\mathrm{\Delta }(g)}\right),& \hfill & \\ \hfill y_2& =\frac{1}{2}((y_1+y_2)-(y_1-y_2))=\frac{1}{2}\left(-A+\sqrt{\mathrm{\Delta }(g)}\right),& \hfill & \end{array}$$

are not in $F$, and this is equivalent to

$$\sqrt{\mathrm{\Delta }(g)}\notin F.$$

Since $\sqrt{\mathrm{\Delta }(g)}\in F(\sqrt{a})$, and $\sqrt{\mathrm{\Delta }(g)}\notin F$,

$$\sqrt{\mathrm{\Delta }(g)}=u+v\sqrt{a},u,v\in F,v\ne 0.$$

Therefore

$$\begin{array}{llll}\hfill u^2& ={\left(\sqrt{\mathrm{\Delta }(g)}-v\sqrt{a}\right)}^2& \hfill & \\ \hfill & =\mathrm{\Delta }(g)+a{v}^2-2v\sqrt{a}\sqrt{\mathrm{\Delta }(g)}& \hfill & \end{array}$$

Since $v\ne 0$, and $\mathrm{char}(F)\ne 2$,

$$\sqrt{a}\sqrt{\mathrm{\Delta }(g)}=\frac{\mathrm{\Delta }(g)+a{v}^2-u^2}{2v}\in F,$$

so

$$a\mathrm{\Delta }(g)\in F^2.$$

$\text{∙}$

Conversely, suppose that $a\mathrm{\Delta }(g)\in F^2$. Here $a\ne 0$ since $F(\sqrt{a})$ is a quadratic extension of $F$. There exists $w\in F$ such that $a\mathrm{\Delta }(g)=w^2$.

We choose $\sqrt{\mathrm{\Delta }(g)}$ such that

$$\sqrt{\mathrm{\Delta }(g)}=\frac{w}{\sqrt{a}}=\frac{w}{a}\sqrt{a}\in F(\sqrt{a}).$$

Then

$$\begin{array}{llll}\hfill y_1& =\frac{1}{2}((y_1+y_2)+(y_1-y_2))=\frac{1}{2}\left(-A-\sqrt{\mathrm{\Delta }(g)}\right),& \hfill & \\ \hfill y_2& =\frac{1}{2}((y_1+y_2)-(y_1-y_2))=\frac{1}{2}\left(-A+\sqrt{\mathrm{\Delta }(g)}\right),& \hfill & \end{array}$$

are in $F(\sqrt{a})$, so $g=(y-y_1)(y-y_2)$ splits completely over $F(\sqrt{a})$.

Finally, if $\mathrm{\Delta }(g)=0$, $g={(y-y_0)}^2$, where $y_0=-A∕2\in F$, splits completely over $F$, a fortiori over $F(\sqrt{a})$.

Conclusion:

Let $g=y^2+\mathit{Ay}+B$ and $F(\sqrt{a})$ a quadratic extension of $F$, with $\mathrm{char}(F)\ne 2$. If $\mathrm{\Delta }(g)=0$, or if $g$ is irreducible over $F$, then

$$g\text{ splits completely over }F(\sqrt{a})⟺a\mathrm{\Delta }(g)\in F^2.$$

(b)

$$\begin{array}{llll}\hfill & (y-({\alpha }_1+{\alpha }_2))(y-({\alpha }_3+{\alpha }_4))& \hfill & \\ \hfill & =y^2-({\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4)y+({\alpha }_1{\alpha }_3+{\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_4)& \hfill & \\ \hfill & =y^2-c_{1}y+({\alpha }_1{\alpha }_2+{\alpha }_1{\alpha }_3+{\alpha }_1{\alpha }_4+{\alpha }_2{\alpha }_3+{\alpha }_2{\alpha }_4+{\alpha }_3{\alpha }_4)-({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4)& \hfill & \\ \hfill & =y^2-c_{1}y+c_2-\beta ,& \hfill & \end{array}$$

so

$$h_1=y^2-c_{1}y+c_2-\beta =(y-({\alpha }_1+{\alpha }_2))(y-({\alpha }_3+{\alpha }_4)).$$

Similarly

$$\begin{array}{llll}\hfill & (y-{\alpha }_1{\alpha }_2)(y-{\alpha }_3{\alpha }_4)& \hfill & \\ \hfill & =y^2-({\alpha }_1{\alpha }_2+{\alpha }_3{\alpha }_4)y+{\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4& \hfill & \\ \hfill & =y^2-\mathit{\beta y}+c_4& \hfill & \end{array}$$

so

$$h_2=y^2-\mathit{\beta y}+c_4=(y-{\alpha }_1{\alpha }_2)(y-{\alpha }_3{\alpha }_4).$$

We have proved that $h_1,h_2$ split completely over $L$. Since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(h_1)=2$, $h_1$ splits over a quadratic extension $F\subset M$, with $M\subset L$. But the unique such quadratic extension is $F(\sqrt{\mathrm{\Delta }(f)})$ (since $\mathrm{Gal}(L∕F)\simeq \mathbb{Z}∕4\mathbb{Z}$ has a unique subgroup of index 2). Therefore $M=F(\sqrt{\mathrm{\Delta }(f)})$, and $h_1$ splits completely over $F(\sqrt{\mathrm{\Delta }(f)})$, and also $h_2$. □