quadratic fields that are not isomorphic

Within this entry, S denotes the set of all squarefreeMathworldPlanetmath integers not equal to 1.


Let m,nS with mn. Then Q(m) and Q(n) are not isomorphicPlanetmathPlanetmathPlanetmath (http://planetmath.org/FieldIsomorphism).


Suppose that (m) and (n) are isomorphic. Let φ:(m)(n) be a field isomorphism. Recall that field homomorphisms fix prime subfields. Thus, for every x, φ(x)=x.

Let a,b with φ(m)=a+bn. Since φ(a)=a and φ is injectivePlanetmathPlanetmath, b0. Also, m=φ(m)=φ((m)2)=(φ(m))2=(a+bn)2=a2+2abn+b2n. If a0, then n=m-a2-b2n2ab, a contradictionMathworldPlanetmathPlanetmath. Thus, a=0. Therefore, m=b2n. Since m is squarefree, b2=1. Hence, m=n, a contradiction. It follows that K and L are not isomorphic. ∎

This yields an obvious corollary:


There are infinitely many distinct quadratic fieldsMathworldPlanetmath.


Note that there are infinitely many elements of S. Moreover, if m and n are distinct elements of S, then (m) and (n) are not isomorphic and thus cannot be equal. ∎

Note that the above corollary could have also been obtained by using the result regarding Galois groupsMathworldPlanetmath of finite abelian extensionsMathworldPlanetmathPlanetmath of (http://planetmath.org/GaloisGroupsOfFiniteAbelianExtensionsOfMathbbQ). On the other hand, using this result to prove the above corollary can be likened to “using a sledgehammer to kill a housefly”.

Title quadratic fields that are not isomorphic
Canonical name QuadraticFieldsThatAreNotIsomorphic
Date of creation 2013-03-22 16:19:44
Last modified on 2013-03-22 16:19:44
Owner Wkbj79 (1863)
Last modified by Wkbj79 (1863)
Numerical id 9
Author Wkbj79 (1863)
Entry type Theorem
Classification msc 11R11