Let $k$ be a field and $K$ be its algebraic closure. Suppose that $k\neq K$ . A quadratic extension $E$ over $k$ is a field $k< E\leq K$ such that $E=k(\alpha)$ for some $\alpha\in K-k$ , where $\alpha^2\in k$ .

If $a=\alpha^2$ , we often write $E=k(\sqrt{a})$ . Every element of $E$ can be written as $r+s\sqrt{a}$ , for some $r,s\in k$ . This representation is unique and we see that $\lbrace 1,\sqrt{a}\rbrace$ is a basis for the vector space $E$ over $k$ . In fact, we have the following

Proposition. If the characteristic of $k$ is not $2$ , then $E$ is a quadratic extension over $k$ iff $\operatorname{dim}(E)=2$ (as a vector space) over $k$ .

Proof. One direction is clear from the above discussion. So suppose $\operatorname{dim}(E)=2$ over $k$ and $\lbrace 1,\beta \rbrace$ is a basis for $E$ over $k$ . Then $\beta^2=r+s\beta$ for some $r,s\in k$ . Set $\alpha=\beta-\frac{s}{2}$ . Then clearly $\alpha\in E-k$ and $\lbrace 1,\alpha\rbrace$ is also a basis for $E$ over $k$ . Furthermore, $\alpha^2= r+\frac{s^2}{4}\in k$ . Thus, $k(\alpha)$ is quadratic extension over $k$ and $[k(\alpha):k]=2$ . But $k(\alpha)$ is a subfield of $E$ . Then $2=[E:k]= [E:k(\alpha)][k(\alpha):k]=2[E:k(\alpha)]$ implies that $[E:k(\alpha)]=1$ and $E=k(\alpha)$ .

In the proposition above, the assumption that $\operatorname{Char}(k)\ne 2$ can not be dropped. If fact, quadratic extensions of $\mathbb{Z}_2$ do not exist, for if $\alpha^2\in \mathbb{Z}_2$ , then $\alpha\in \mathbb{Z}_2$ .

For the rest of the discussion, we assume that $\operatorname{Char}(k)\ne 2$ .

Pick any element $\beta=r+s\sqrt{a}$ in $E-k$ . Then $s\neq 0$ and $(\beta - r)^2=s^2a\in k$ . So $\beta$ is a root of the irreducible polynomial $m(x)=x^2-2rx+(r^2-s^2a)$ in $k[x]$ . If we define $\overline{\beta}$ to be $r-s\sqrt{a}$ , then $\overline{\beta}$ is the other root of $m(x)$ , clearly also in $E-k$ . This implies that the minimal polynomial of every element in $E$ has degree at most 2, and splits into linear factors in $E[x]$ .

Since $\operatorname{Char}(k)\ne 2$ , $\beta\neq\overline{\beta}$ are two distinct roots of $m(x)$ . This shows that $k(\sqrt{a})$ is separable over $k$ .

Now, let $f(x)$ be any irreducible polynomial over $k$ which has a root $\beta$ in $E$ . Then the minimal polynomial $m(x)$ of $\beta$ in $k[x]$ must divide $f$ . But because $f$ is irreducible, $m=f$ . This shows that $k(\sqrt{a})$ is normal over $k$ . Since $k(\sqrt{a})$ is both separable and normal over $k$ , it is a Galois extension over $k$ .

Let $\phi$ be an automorphism of $E=k(\sqrt{a})$ fixing $k$ . Then $\phi(\sqrt{a})$ is easily seen to be a root of the minimal polynomial of $\sqrt{a}$ . As a result, either $\phi=1$ on $E$ or $\phi$ is the involution that maps each $\beta$ to $\overline{\beta}$ . We have just proved

Theorem. Suppose $\operatorname{Char}(k)\neq 2$ . Any quadratic extension of $k$ is Galois over $k$ , whose Galois group is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ .

Remark. A quadratic extension (of a field) is also known in the literature as a $2$ -extension, a special case of a p-extension, when $p=2$ .