# 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 $\mathrm{dim}(E)=2$ (as a vector space) over $k$.

###### Proof.

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