quadratic extension

Let k be a field and K be its algebraic closureMathworldPlanetmath. Suppose that kK. A quadratic extension E over k is a field k<EK such that E=k(α) for some αK-k, where α2k.

If a=α2, we often write E=k(a). Every element of E can be written as r+sa, for some r,sk. This representation is unique and we see that {1,a} is a basis for the vector spaceMathworldPlanetmath E over k. In fact, we have the following

PropositionPlanetmathPlanetmathPlanetmath. If the characteristicPlanetmathPlanetmath of k is not 2, then E is a quadratic extension over k iff dim(E)=2 (as a vector space) over k.


One direction is clear from the above discussion. So suppose dim(E)=2 over k and {1,β} is a basis for E over k. Then β2=r+sβ for some r,sk. Set α=β-s2. Then clearly αE-k and {1,α} is also a basis for E over k. Furthermore, α2=r+s24k. Thus, k(α) is quadratic extension over k and [k(α):k]=2. But k(α) is a subfieldMathworldPlanetmath of E. Then 2=[E:k]=[E:k(α)][k(α):k]=2[E:k(α)] implies that [E:k(α)]=1 and E=k(α). ∎

In the proposition above, the assumptionPlanetmathPlanetmath that Char(k)2 can not be dropped. If fact, quadratic extensions of 2 do not exist, for if α22, then α2.

For the rest of the discussion, we assume that Char(k)2.

Pick any element β=r+sa in E-k. Then s0 and (β-r)2=s2ak. So β is a root of the irreducible polynomial m(x)=x2-2rx+(r2-s2a) in k[x]. If we define β¯ to be r-sa, then β¯ is the other root of m(x), clearly also in E-k. This implies that the minimal polynomialPlanetmathPlanetmath of every element in E has degree at most 2, and splits into linear factors in E[x].

Since Char(k)2, ββ¯ are two distinct roots of m(x). This shows that k(a) is separablePlanetmathPlanetmath over k.

Now, let f(x) be any irreducible polynomial over k which has a root β in E. Then the minimal polynomial m(x) of β in k[x] must divide f. But because f is irreducible, m=f. This shows that k(a) is normal over k. Since k(a) is both separable and normal over k, it is a Galois extensionMathworldPlanetmath over k.

Let ϕ be an automorphismPlanetmathPlanetmathPlanetmathPlanetmath of E=k(a) fixing k. Then ϕ(a) is easily seen to be a root of the minimal polynomial of a. As a result, either ϕ=1 on E or ϕ is the involutionPlanetmathPlanetmath that maps each β to β¯. We have just proved

Theorem. Suppose Char(k)2. Any quadratic extension of k is Galois over k, whose Galois groupMathworldPlanetmath is isomorphic to /2.

Remark. A quadratic extension (of a field) is also known in the literature as a 2-extensionPlanetmathPlanetmathPlanetmathPlanetmath, 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