biquadratic field


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

Let k=mn(gcd(m,n))2. It can easily be verified that kS, km, and kn. Since k=mngcd(m,n)(m,n), the three distinct quadratic subfieldsMathworldPlanetmath of (m,n) are (m), (n), and (k). Note that (k)=(mn).

Of the three cyclotomic fieldsMathworldPlanetmath of degree (http://planetmath.org/ExtensionField) four over , (ω8) and (ω12) are biquadratic fields. The quadratic subfields of (ω8) are (2), (-1), and (-2); the quadratic subfields of (ω12) are (3), (-1), and (-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 idealMathworldPlanetmathPlanetmath P over p.) This occurs for biquadratic fields (m,n) in which exactly two of m, n, and k are equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath (http://planetmath.org/CongruencesMathworldPlanetmathPlanetmathPlanetmathPlanetmath) to 2mod4 and the other is to 3mod4. For example, in (ω8)=(2,-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=(2,-3) is useful for demonstrating that a subring of a principal ideal domainMathworldPlanetmath need not be a principal ideal domain. It can easily be verified that 𝒪K (the ring of integersMathworldPlanetmath of K) is a principal ideal domain, but [-6], which is a subring of 𝒪K, is not a principal ideal domain. Also, biquadratic fields of the form L=(m,n) with m and n distinct elements of S such that m1mod3 and n1mod3 are useful for demonstrating that rings of integers need not have power bases over (http://planetmath.org/PowerBasisOverMathbbZ). Note that 3 splits completely in both (m) and (n) and thus in L. Therefore, 3𝒪L=P1P2P3P4 for distinct prime ideals (http://planetmath.org/PrimeIdeal) P1, P2, P3, and P4 of 𝒪L. Now suppose 𝒪L=[α] for some αL. Then L=(α), and the minimal polynomial f for α over has degree 4. This yields that f, considered as a polynomialMathworldPlanetmathPlanetmathPlanetmath over 𝔽3, is supposed to factor into four distinct monic polynomialsMathworldPlanetmath of degree 1, which is a contradictionMathworldPlanetmathPlanetmath.

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