Exercise 7.1.12

Proof. Let $F\subset L$ be an extension of degree 2, where $F$ has characteristic $\ne 2$. Then $[L:F]=2,F\subset L,F\ne L$.

(a)

Let $\alpha \in L\setminus F$. Then $(1,\alpha )$ is a linearly independent list, otherwise $\alpha \in F$. As ${\dim <!--\; FUNCTION\; APPLICATION\; -->}_F(L)=2$, $(1,\alpha )$ is a basis of of the $F$-vector space $L$.

Therefore there exists a pair $(a,b)\in F^2$ such that ${\alpha }^2=\mathit{a\alpha }+b$, so $\alpha$ is a root of the polynomial $f=x^2-\mathit{ax}-b\in F[x]$.

$$(x-\alpha )(x-(a-\alpha ))=x^2-\mathit{ax}+\alpha (a-\alpha )=x^2-\mathit{ax}-b=f.$$

The roots of $f$ are so $\alpha$ and $\beta =a-\alpha$, both in $L$.

As $\alpha \notin F$, $1<[F(\alpha ):F]\le 2$, thus $[F(\alpha ):F]=2=[L:F]$ with $F[\alpha ]\subset L$, therefore $L=F(\alpha )$.

The polynomial $f\in F[x]$ is irreducible over $F$ since $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)=2$ and the roots of $f$ are $\alpha \notin F,a-\alpha \notin F$. So $f$ is the minimal polynomial of $\alpha$ over $F$.

(b)

The roots of $f$, minimal polynomial of $\alpha$ over $F$, are $\alpha ,\beta$, which are distinct, otherwise $\alpha =a-\alpha$, and then $\alpha =a∕2\in F$ (the characteristic is not equal to 2), which is false. The minimal polynomial of $\alpha$ over $F$ is so separable.

(c)

As $\beta =a-\alpha ,a\in F$, $\beta \in F(\alpha )$, thus $L=F(\alpha )=F(\alpha ,\beta )$ is the splitting field of the separable polynomial $f\in F[x]$. Therefore, by Theorem 7.1.1, $F\subset L$ is a Galois extension.

$f$ being irreducible, there exists (Prop. 5.1.8) an isomorphism $\sigma :L\to L$ such that $\sigma (\alpha )=\beta$ and $\sigma$ is the identity on $F$, so $\sigma \in \mathrm{Gal}(L∕F)$.

(Explicitly, $\sigma :u+\mathit{v\alpha }\mapsto u+\mathit{v\beta },u,v\in F$: we can verify directly that it is an isomorphism.)

Every $\tau \in \mathrm{Gal}(L∕F)$ sends the root $\alpha$ of $f\in F[x]$ on a root of $f$, so $\tau (\alpha )=\alpha =1_K(\alpha )$ or $\tau (\alpha )=\beta =\sigma (\alpha )$. As $L=F(\alpha )$, this $F$-automorphisme is uniquely determined by the image of $\alpha$. Thus $\tau =\sigma$ or $\tau =1_K=e$. Moreover $\sigma \ne e$, otherwise $\sigma (\alpha )=\alpha$, so $\beta =\alpha$, which is false by part (b). Consequently $G=\{e,\sigma \}$.

Every group of order 2 is isomorphic to $\mathbb{Z}∕2\mathbb{Z}$, thus

$$G=\{e,\sigma \}\simeq \mathbb{Z}∕2\mathbb{Z}.$$

(d)

As the characteristic is not 2, $$0={\alpha }^2-\mathit{a\alpha }-b={\left(\alpha -\frac{a}{2}\right)}^2-\frac{a^2}{4}-b.$$

Therefore $\gamma =\alpha -\frac{a}{2}$ satisfies ${\gamma }^2=\frac{a^2+4b^2}{4}\in F$.

As $\gamma =\alpha -\frac{a}{2}$ with $a\in F$, $F(\gamma )=F(\alpha )=L$. Write $c={\gamma }^2\in F$, and $\sqrt{c}=\gamma$, then

$$L=F(\gamma ),{\gamma }^2\in F,L=F(\sqrt{c}),c\in F.$$