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# quadratic extension

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 $\{1,\sqrt{a}\}$ 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 $\{1,\beta\}$ 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 $\{1,\alpha\}$ 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 $\operatorname{Char}(k)\neq 2$ 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 $\operatorname{Char}(k)\neq 2$.

Pick any element $\beta=r+s\sqrt{a}$ in $E-k$. Then $s\neq 0$ and $(\beta-r)^{2}=s^{2}a\in k$. So $\beta$ is a root of the irreducible polynomial $m(x)=x^{2}-2rx+(r^{2}-s^{2}a)$ 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 $\operatorname{Char}(k)\neq 2$, $\beta\neq\overline{\beta}$ 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 $p=2$.

## Mathematics Subject Classification

12F05*no label found*12F10

*no label found*

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