Exercise 13.1.5

Proof. Let $g=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)$, where ${\alpha }_1,{\alpha }_2,{\alpha }_3$ lie in some splitting field of $F$, and ${\alpha }_1\in F$.

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If $g$ splits completely over $F$, then ${\alpha }_1,{\alpha }_2,{\alpha }_3$ lie in $F$, therefore

$\delta =({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)({\alpha }_2-{\alpha }_3)\in F$, so $\mathrm{\Delta }(g)={\delta }^2\in F^2$.

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Conversely, suppose that $\mathrm{\Delta }(g)\in F^2$. Then $\mathrm{\Delta }(g)=a^2,a\in F$, so $\delta =\pm a\in F$. Since ${\alpha }_1\in F$, the Euclidean division of $g(x)$ by $x-{\alpha }_1\in F[x]$ gives $$g(x)=(x-{\alpha }_1)(x^2+\mathit{px}+q),p,q\in F.$$

Then $x^2+\mathit{px}+q=(x-{\alpha }_2)(x-{\alpha }_3)$, hence ${\alpha }_2+{\alpha }_3=-p\in F,{\alpha }_2{\alpha }_3=q\in F$, and

$$({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)={\alpha }_1^2+p{\alpha }_1+q\in F.$$

If ${\alpha }_1={\alpha }_2$, then ${\alpha }_3=-p-{\alpha }_2=-p-{\alpha }_1\in F$, so $g$ splits completely over $F$, and similarly the same conclusion is true if ${\alpha }_1={\alpha }_3$.

In the remaining case, $({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)\ne 0$, so

$${\alpha }_2-{\alpha }_3=\delta {[({\alpha }_1-{\alpha }_2)({\alpha }_1-{\alpha }_3)]}^{-1}\in F.$$

Since ${\alpha }_2+{\alpha }_3\in F$, and ${\alpha }_2-{\alpha }_3\in F$, and since the characteristic of $F$ is not 2,

$${\alpha }_2=\frac{1}{2}[({\alpha }_2+{\alpha }_3)+({\alpha }_2-{\alpha }_3)]\in F,{\alpha }_3=\frac{1}{2}[({\alpha }_2+{\alpha }_3)-({\alpha }_2-{\alpha }_3)]\in F.$$

Therefore $g=(x-{\alpha }_1)(x-{\alpha }_2)(x-{\alpha }_3)$ splits completely over $F$.