# biquadratic field

A biquadratic field (or biquadratic number field) is a biquadratic extension of $\mathbb{Q}$. To discuss these fields more easily, the set $S$ will be defined to be the set of all squarefree^{} integers not equal to $1$. Thus, any biquadratic field is of the form $\mathbb{Q}(\sqrt{m},\sqrt{n})$ for distinct elements $m$ and $n$ of $S$.

Let $k={\displaystyle \frac{mn}{{(\mathrm{gcd}(m,n))}^{2}}}$. It can easily be verified that $k\in S$, $k\ne m$, and $k\ne n$. Since $\sqrt{k}={\displaystyle \frac{\sqrt{mn}}{\mathrm{gcd}(m,n)}}\in \mathbb{Q}(\sqrt{m},\sqrt{n})$, the three distinct quadratic subfields^{} of $\mathbb{Q}(\sqrt{m},\sqrt{n})$ are $\mathbb{Q}(\sqrt{m})$, $\mathbb{Q}(\sqrt{n})$, and $\mathbb{Q}(\sqrt{k})$. Note that $\mathbb{Q}(\sqrt{k})=\mathbb{Q}(\sqrt{mn})$.

Of the three cyclotomic fields^{} of degree (http://planetmath.org/ExtensionField) four over $\mathbb{Q}$, $\mathbb{Q}({\omega}_{8})$ and $\mathbb{Q}({\omega}_{12})$ are biquadratic fields. The quadratic subfields of $\mathbb{Q}({\omega}_{8})$ are $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{-1})$, and $\mathbb{Q}(\sqrt{-2})$; the quadratic subfields of $\mathbb{Q}({\omega}_{12})$ are $\mathbb{Q}(\sqrt{3})$, $\mathbb{Q}(\sqrt{-1})$, and $\mathbb{Q}(\sqrt{-3})$.

Note that the only rational prime $p$ for which $e(P|p)=4$ is possible in a biquadratic field is $p=2$. (The notation $e(P|p)$ refers to the ramification index of the prime ideal^{} $P$ over $p$.) This occurs for biquadratic fields $\mathbb{Q}(\sqrt{m},\sqrt{n})$ in which exactly two of $m$, $n$, and $k$ are equivalent^{} (http://planetmath.org/Congruences^{}) to $2\mathrm{mod}4$ and the other is to $3\mathrm{mod}4$. For example, in $\mathbb{Q}({\omega}_{8})=\mathbb{Q}(\sqrt{2},\sqrt{-1})$, we have that $e(P|2)=4$.

Certain biquadratic fields provide excellent counterexamples to statements that some people might think to be true. For example, the biquadratic field $K=\mathbb{Q}(\sqrt{2},\sqrt{-3})$ is useful for demonstrating that a subring of a principal ideal domain^{} need not be a principal ideal domain. It can easily be verified that ${\mathcal{O}}_{K}$ (the ring of integers^{} of $K$) is a principal ideal domain, but $\mathbb{Z}[\sqrt{-6}]$, which is a subring of ${\mathcal{O}}_{K}$, is not a principal ideal domain. Also, biquadratic fields of the form $L=\mathbb{Q}(\sqrt{m},\sqrt{n})$ with $m$ and $n$ distinct elements of $S$ such that $m\equiv 1\mathrm{mod}3$ and $n\equiv 1\mathrm{mod}3$ are useful for demonstrating that rings of integers need not have power bases over $\mathbb{Z}$ (http://planetmath.org/PowerBasisOverMathbbZ). Note that $3$ splits completely in both $\mathbb{Q}(\sqrt{m})$ and $\mathbb{Q}(\sqrt{n})$ and thus in $L$. Therefore, $3{\mathcal{O}}_{L}={P}_{1}{P}_{2}{P}_{3}{P}_{4}$ for distinct prime ideals (http://planetmath.org/PrimeIdeal) ${P}_{1}$, ${P}_{2}$, ${P}_{3}$, and ${P}_{4}$ of ${\mathcal{O}}_{L}$. Now suppose ${\mathcal{O}}_{L}=\mathbb{Z}[\alpha ]$ for some $\alpha \in L$. Then $L=\mathbb{Q}(\alpha )$, and the minimal polynomial $f$ for $\alpha $ over $\mathbb{Q}$ has degree $4$. This yields that $f$, considered as a polynomial^{} over ${\mathbb{F}}_{3}$, is supposed to factor into four distinct monic polynomials^{} of degree $1$, which is a contradiction^{}.

Title | biquadratic field |
---|---|

Canonical name | BiquadraticField |

Date of creation | 2013-03-22 15:56:24 |

Last modified on | 2013-03-22 15:56:24 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 16 |

Author | Wkbj79 (1863) |

Entry type | Definition |

Classification | msc 11R16 |

Synonym | biquadratic number field |

Related topic | BiquadraticExtension |

Related topic | BiquadraticEquation2 |