quadratic extension

Let $k$ be a field and $K$ be its algebraic closure. Suppose that $k\ne K$. A quadratic extension $E$ over $k$ is a field 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 $\dim (E)=2$ (as a vector space) over $k$.

Proof.

One direction is clear from the above discussion. So suppose $\dim (E)=2$ 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 $\mathrm{Char}(k)\ne 2$ 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 $\mathrm{Char}(k)\ne 2$.

Pick any element $\beta =r+s\sqrt{a}$ in $E-k$. Then $s\ne 0$ 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 $\mathrm{Char}(k)\ne 2$, $\beta \ne \overline{\beta }$ 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 $\varphi$ be an automorphism of $E=k(\sqrt{a})$ fixing $k$. Then $\varphi (\sqrt{a})$ is easily seen to be a root of the minimal polynomial of $\sqrt{a}$. As a result, either $\varphi =1$ on $E$ or $\varphi$ is the involution that maps each $\beta$ to $\overline{\beta }$. We have just proved

Theorem. Suppose $\mathrm{Char}(k)\ne 2$. 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 $\mathrm{2}$-extension, a special case of a p-extension, when $p=2$.

Title	quadratic extension
Canonical name	QuadraticExtension
Date of creation	2013-03-22 15:42:34
Last modified on	2013-03-22 15:42:34
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	20
Author	CWoo (3771)
Entry type	Definition
Classification	msc 12F05
Classification	msc 12F10
Synonym	$2$-extension
Related topic	PExtension